Current Electricity MCQs with Answers – Part 7 (NEET Physics)

Explanation: \(\textbf{Determine the initial equivalent internal resistance:}\) The two identical internal resistances are initially parallel: \[ r_{\text{eq},i}=\frac{(2.0)(2.0)}{2.0+2.0} \] \[ r_{\text{eq},i}=1.0\,\Omega \] The equivalent EMF remains \[ \mathcal E_{\text{eq}}=12\,\text{V} \] \(\textbf{Calculate the initial load current and voltage:}\) \[ I_i=\frac{\mathcal E_{\text{eq}}}{R+r_{\text{eq},i}} \] \[ I_i=\frac{12}{3.0+1.0} \] \[ I_i=3.0\,\text{A} \] Therefore, \[ V_i=I_iR \] \[ V_i=(3.0)(3.0) \] \[ V_i=9.0\,\text{V} \] \(\textbf{Recalculate the equivalent resistance after ageing:}\) The internal resistances are now \(2.0\,\Omega\) and \(4.0\,\Omega\): \[ r_{\text{eq},f}=\frac{(2.0)(4.0)}{2.0+4.0} \] \[ r_{\text{eq},f}=\frac43\,\Omega \] Their EMFs remain identical, so the equivalent EMF is still \(12\,\text{V}\). \(\textbf{Find the new load current:}\) \[ I_f=\frac{12}{3.0+4/3} \] \[ I_f=\frac{36}{13}\,\text{A} \] Thus, \[ V_f=I_fR \] \[ V_f=\frac{108}{13}\,\text{V} \] \(\textbf{Determine individual cell currents:}\) Each real cell has the same terminal voltage. For P, \[ I_P=\frac{12-V_f}{2.0} \] \[ I_P=\frac{12-108/13}{2.0} \] \[ I_P=\frac{24}{13}\,\text{A} \] For Q, \[ I_Q=\frac{12-V_f}{4.0} \] \[ I_Q=\frac{12}{13}\,\text{A} \] Therefore, \[ I_Q:I_P=1:2 \] \(\textbf{Compare external-load powers:}\) The external resistance is unchanged. Hence, \[ \frac{P_f}{P_i}=\left(\frac{V_f}{V_i}\right)^2 \] \[ \frac{P_f}{P_i}=\left(\frac{108/13}{9.0}\right)^2 \] \[ \frac{P_f}{P_i}=\left(\frac{12}{13}\right)^2 \] \[ \frac{P_f}{P_i}=\frac{144}{169} \] \(\textbf{Physical interpretation:}\) Cells with equal EMFs share current in inverse proportion to their internal resistances. Increasing Q's internal resistance reduces its contribution to the external current. The total equivalent internal resistance increases, so the load terminal voltage decreases. The external power decreases even though neither cell's EMF changes. The healthy cell now supplies twice the current supplied by the aged cell. \(\textbf{Final answer:}\) \[ \boxed{I_Q:I_P=1:2,\qquad\frac{P_f}{P_i}=\frac{144}{169}} \]

Explanation: \(\textbf{Determine the total load current:}\) The external resistance is \[ R=3.0\,\Omega \] The voltmeter reads \[ V=9.0\,\text{V} \] Therefore, \[ I_{\text{load}}=\frac VR \] \[ I_{\text{load}}=\frac{9.0}{3.0} \] \[ I_{\text{load}}=3.0\,\text{A} \] \(\textbf{Calculate the currents from the known cells:}\) Each cell supplies current according to \[ I=\frac{\mathcal E-V}{r} \] For the \(2.0\,\Omega\) cell, \[ I_1=\frac{12-9.0}{2.0} \] \[ I_1=1.50\,\text{A} \] For the \(3.0\,\Omega\) cell, \[ I_2=\frac{12-9.0}{3.0} \] \[ I_2=1.0\,\text{A} \] \(\textbf{Use the junction rule to find the third current:}\) \[ I_1+I_2+I_3=I_{\text{load}} \] \[ 1.50+1.0+I_3=3.0 \] \[ I_3=0.50\,\text{A} \] \(\textbf{Recover the unknown internal resistance:}\) \[ r=\frac{\mathcal E-V}{I_3} \] \[ r=\frac{12-9.0}{0.50} \] \[ r=6.0\,\Omega \] \(\textbf{Determine total chemical-energy conversion rate:}\) The three cells have identical EMFs. Therefore, \[ P_{\text{chemical}}=\mathcal E(I_1+I_2+I_3) \] \[ P_{\text{chemical}}=(12)(3.0) \] \[ P_{\text{chemical}}=36\,\text{W} \] \(\textbf{Determine internal heating in all three cells:}\) \[ P_{\text{internal}}=I_1^2r_1+I_2^2r_2+I_3^2r_3 \] Substituting, \[ P_{\text{internal}}=(1.50)^2(2.0)+(1.0)^2(3.0)+(0.50)^2(6.0) \] \[ P_{\text{internal}}=4.50+3.0+1.50 \] \[ P_{\text{internal}}=9.0\,\text{W} \] \(\textbf{Find the required fraction:}\) \[ f=\frac{P_{\text{internal}}}{P_{\text{chemical}}} \] \[ f=\frac{9.0}{36} \] \[ f=\frac14 \] \(\textbf{Verify energy conservation:}\) The load receives \[ P_{\text{load}}=VI_{\text{load}} \] \[ P_{\text{load}}=(9.0)(3.0)=27\,\text{W} \] Adding internal heating, \[ 27+9.0=36\,\text{W} \] This equals the total chemical-energy conversion rate. The cell with the greatest internal resistance supplies the smallest current, but its internal heating is not necessarily the largest. The individual branch currents must be calculated before their losses are combined. \(\textbf{Final answer:}\) \[ \boxed{r=6.0\,\Omega,\qquad f=\frac14} \]

Explanation: \(\textbf{Determine the equivalent source parameters:}\) The two cell currents are \[ I_P=\frac{16-V}{2.0} \] and \[ I_Q=\frac{8.0-V}{2.0} \] Their combined current is \[ I=12-V \] Rearranging, \[ V=12-I(1.0) \] Thus, the parallel-cell arrangement is equivalent at its external terminals to \[ \mathcal E_{\text{eq}}=12\,\text{V} \] with \[ r_{\text{eq}}=1.0\,\Omega \] \(\textbf{Determine the initial load voltage:}\) The original load is \(4.0\,\Omega\). Therefore, \[ I_i=\frac{12}{4.0+1.0} \] \[ I_i=\frac{12}{5}\,\text{A} \] The terminal voltage is \[ V_i=I_i(4.0) \] \[ V_i=\frac{48}{5}\,\text{V} \] \(\textbf{Identify Q's initial operating direction:}\) \[ I_{Q,i}=\frac{8.0-48/5}{2.0} \] \[ I_{Q,i}=-\frac45\,\text{A} \] Initially, Q charges at \(0.80\,\text{A}\). \(\textbf{Determine the modified external resistance:}\) The two \(4.0\,\Omega\) external resistors are connected in parallel: \[ R_f=\frac{(4.0)(4.0)}{4.0+4.0} \] \[ R_f=2.0\,\Omega \] \(\textbf{Determine the new terminal voltage:}\) \[ I_f=\frac{12}{2.0+1.0} \] \[ I_f=4.0\,\text{A} \] Therefore, \[ V_f=I_fR_f \] \[ V_f=(4.0)(2.0) \] \[ V_f=8.0\,\text{V} \] \(\textbf{Determine Q's final current:}\) \[ I_{Q,f}=\frac{8.0-8.0}{2.0} \] \[ I_{Q,f}=0\,\text{A} \] The changed load lowers the common terminal voltage exactly to Q's EMF. \(\textbf{Compare total external-load powers:}\) Initially, \[ P_i=\frac{V_i^2}{4.0} \] \[ P_i=\frac{(48/5)^2}{4.0} \] \[ P_i=\frac{576}{25}\,\text{W} \] Finally, \[ P_f=\frac{V_f^2}{2.0} \] \[ P_f=\frac{(8.0)^2}{2.0} \] \[ P_f=32\,\text{W} \] Hence, \[ \frac{P_f}{P_i}=\frac{32}{576/25} \] \[ \frac{P_f}{P_i}=\frac{25}{18} \] \(\textbf{Physical interpretation:}\) Initially, P supplies the load and also charges Q. Adding a parallel external resistor increases the current demanded by the load. The common terminal voltage falls, reducing the charging current through Q to zero. At the same time, the total load power increases because the effective load resistance decreases sufficiently. A change in external loading can alter a cell's operating state without any change in the cell's polarity or EMF. \(\textbf{Final answer:}\) \[ \boxed{I_{Q,f}=0\,\text{A},\qquad\frac{P_f}{P_i}=\frac{25}{18}} \]

Explanation: \(\textbf{Compare the source arrangements through their equivalent parameters:}\) For each arrangement, calculate the equivalent EMF and internal resistance. The load power is \[ P_R=\left(\frac{\mathcal E_{\text{eq}}}{R+r_{\text{eq}}}\right)^2R \] \(\textbf{Arrangement P — Four cells in series:}\) The EMFs add: \[ \mathcal E_P=4(3.0)=12\,\text{V} \] The internal resistances also add: \[ r_P=4(1.0)=4.0\,\Omega \] The load current is \[ I_P=\frac{12}{1.0+4.0} \] \[ I_P=\frac{12}{5}\,\text{A} \] Therefore, \[ P_{R,P}=\left(\frac{12}{5}\right)^2(1.0) \] \[ P_{R,P}=\frac{144}{25}\,\text{W} \] \(\textbf{Arrangement Q — Two series pairs connected in parallel:}\) Each pair has EMF \[ \mathcal E_{\text{pair}}=2(3.0)=6.0\,\text{V} \] and internal resistance \[ r_{\text{pair}}=2(1.0)=2.0\,\Omega \] Two identical pairs in parallel give \[ \mathcal E_Q=6.0\,\text{V} \] \[ r_Q=\frac{2.0}{2}=1.0\,\Omega \] The external current is \[ I_Q=\frac{6.0}{1.0+1.0} \] \[ I_Q=3.0\,\text{A} \] Thus, \[ P_{R,Q}=(3.0)^2(1.0) \] \[ P_{R,Q}=9.0\,\text{W} \] \(\textbf{Arrangement R — Four cells directly in parallel:}\) The equivalent EMF is \[ \mathcal E_R=3.0\,\text{V} \] The equivalent internal resistance is \[ r_R=\frac{1.0}{4} \] \[ r_R=0.25\,\Omega \] The load current is \[ I_R=\frac{3.0}{1.0+0.25} \] \[ I_R=\frac{12}{5}\,\text{A} \] Therefore, \[ P_{R,R}=\frac{144}{25}\,\text{W} \] \(\textbf{Select the maximum-power arrangement:}\) The external powers are \[ P_{R,P}=\frac{144}{25}\,\text{W} \] \[ P_{R,Q}=9.0\,\text{W} \] \[ P_{R,R}=\frac{144}{25}\,\text{W} \] Arrangement Q gives the greatest external power. \(\textbf{Determine individual cell currents in arrangement Q:}\) The two parallel branches are identical. They share the total \(3.0\,\text{A}\) equally: \[ I_{\text{branch}}=\frac{3.0}{2} \] \[ I_{\text{branch}}=1.50\,\text{A} \] Each cell in a series pair carries this current. \(\textbf{Calculate total internal heating:}\) There are four cells, each with internal resistance \(1.0\,\Omega\). Therefore, \[ P_{\text{internal}}=4I_{\text{branch}}^2r \] \[ P_{\text{internal}}=4(1.50)^2(1.0) \] \[ P_{\text{internal}}=9.0\,\text{W} \] \(\textbf{Physical interpretation:}\) Increasing the number of cells in series raises both the available EMF and the total internal resistance. Increasing the number of identical parallel branches reduces the effective internal resistance. For the stated load, the two-series-pairs arrangement provides the most effective combination. Its equivalent internal resistance equals the external resistance, producing equal internal and external heating rates at this operating point. \(\textbf{Final answer:}\) \[ \text{Arrangement Q},\qquad P_{\text{internal}}=9.0\,\text{W} \]

Explanation: \(\textbf{Construct the equivalent parallel-cell source:}\) Let the common terminal voltage be V. The algebraic currents supplied by the two cells are \[ I_P=\frac{18-V}{1.0} \] \[ I_Q=\frac{6.0-V}{3.0} \] The combined external current is \[ I=I_P+I_Q \] Therefore, \[ I=18-V+\frac{6.0-V}{3.0} \] Multiplying by three, \[ 3I=60-4V \] Rearranging, \[ V=15-\frac34I \] Thus, the equivalent source has \[ \mathcal E_{\text{eq}}=15\,\text{V} \] and \[ r_{\text{eq}}=\frac34\,\Omega \] \(\textbf{Apply the maximum external-power condition:}\) For a source with constant equivalent EMF and internal resistance, the external-load power is greatest when \[ R=r_{\text{eq}} \] Hence, \[ R_{\text{opt}}=\frac34\,\Omega \] \(\textbf{Determine the maximum-power current:}\) \[ I_{\text{opt}}=\frac{\mathcal E_{\text{eq}}}{R_{\text{opt}}+r_{\text{eq}}} \] \[ I_{\text{opt}}=\frac{15}{3/4+3/4} \] \[ I_{\text{opt}}=10\,\text{A} \] \(\textbf{Calculate the maximum load power:}\) \[ P_{\max}=I_{\text{opt}}^2R_{\text{opt}} \] \[ P_{\max}=(10)^2\left(\frac34\right) \] \[ P_{\max}=75\,\text{W} \] \(\textbf{Determine the common terminal voltage:}\) \[ V=I_{\text{opt}}R_{\text{opt}} \] \[ V=(10)\left(\frac34\right) \] \[ V=7.5\,\text{V} \] \(\textbf{Identify Q's current direction:}\) \[ I_Q=\frac{6.0-7.5}{3.0} \] \[ I_Q=-0.50\,\text{A} \] The actual current enters Q's positive terminal. Therefore, Q is charging. \(\textbf{Calculate Q's chemical-energy storage rate:}\) \[ P_{\text{chemical},Q}=\mathcal E_QI_{\text{ch},Q} \] \[ P_{\text{chemical},Q}=(6.0)(0.50) \] \[ P_{\text{chemical},Q}=3.0\,\text{W} \] \(\textbf{Check current conservation:}\) Cell P supplies \[ I_P=\frac{18-7.5}{1.0} \] \[ I_P=10.5\,\text{A} \] After subtracting the \(0.50\,\text{A}\) entering Q, \[ 10.5-0.50=10\,\text{A} \] This equals the external-load current. \(\textbf{Physical interpretation:}\) The equivalent source determines the maximum possible external power, but it does not reveal every cell's operating state by itself. Individual branch relations must be used after the common terminal voltage is found. At this maximum-power condition, the higher-EMF cell supplies the load and also charges the lower-EMF cell. Maximum external power therefore does not require every parallel cell to discharge. \(\textbf{Final answer:}\) \[ \boxed{P_{\max}=75\,\text{W},\qquad P_{\text{chemical},Q}=3.0\,\text{W}} \]

Explanation: \(\textbf{Apply current conservation with no external load:}\) Let the common positive-to-negative terminal voltage be V. The algebraic currents supplied by the three cells are \[ I_P=\frac{18-V}{2.0} \] \[ I_Q=\frac{12-V}{3.0} \] \[ I_S=\frac{6.0-V}{6.0} \] There is no external current path. Therefore, \[ I_P+I_Q+I_S=0 \] \(\textbf{Solve for the common terminal voltage:}\) Substituting, \[ \frac{18-V}{2.0}+\frac{12-V}{3.0}+\frac{6.0-V}{6.0}=0 \] Multiplying by six, \[ 3(18-V)+2(12-V)+(6.0-V)=0 \] \[ 54-3V+24-2V+6.0-V=0 \] \[ 84=6V \] \[ V=14\,\text{V} \] \(\textbf{Determine the current supplied by P:}\) \[ I_P=\frac{18-14}{2.0} \] \[ I_P=2.0\,\text{A} \] Thus, P discharges. \(\textbf{Determine Q's charging current:}\) \[ I_Q=\frac{12-14}{3.0} \] \[ I_Q=-\frac23\,\text{A} \] The negative sign indicates charging current. Its magnitude is \(2/3\,\text{A}\). \(\textbf{Determine S's charging current:}\) \[ I_S=\frac{6.0-14}{6.0} \] \[ I_S=-\frac43\,\text{A} \] S also charges. \(\textbf{Check the circulating-current balance:}\) \[ I_P+I_Q+I_S=2.0-\frac23-\frac43 \] \[ I_P+I_Q+I_S=0 \] No current leaves the parallel-cell assembly through an external load. \(\textbf{Calculate chemical-energy storage in Q and S:}\) For Q, \[ P_Q=\mathcal E_Q|I_Q| \] \[ P_Q=(12)\left(\frac23\right) \] \[ P_Q=8.0\,\text{W} \] For S, \[ P_S=\mathcal E_S|I_S| \] \[ P_S=(6.0)\left(\frac43\right) \] \[ P_S=8.0\,\text{W} \] Therefore, \[ P_{\text{stored}}=P_Q+P_S \] \[ P_{\text{stored}}=16\,\text{W} \] \(\textbf{Verify the energy balance:}\) Cell P generates \[ P_{\text{generated}}=(18)(2.0)=36\,\text{W} \] Internal heating in the three cells is \[ P_{\text{internal}}=(2.0)^2(2.0)+\left(\frac23\right)^2(3.0)+\left(\frac43\right)^2(6.0) \] \[ P_{\text{internal}}=8.0+\frac43+\frac{32}{3} \] \[ P_{\text{internal}}=20\,\text{W} \] Adding chemical storage, \[ 20+16=36\,\text{W} \] Energy conservation is satisfied. \(\textbf{Physical interpretation:}\) An open external circuit does not eliminate the closed paths formed by unequal cells connected in parallel. The highest-EMF cell supplies circulating current. The two lower-EMF cells receive charging current. The common terminal voltage lies between the smallest and largest cell EMFs. Zero external current is therefore compatible with nonzero internal heating and chemical-energy transfer. \(\textbf{Final answer:}\) \[ \boxed{V=14\,\text{V},\qquad P_{\text{stored}}=16\,\text{W}} \]

Explanation: \(\textbf{Determine the original terminal voltage:}\) Let the common terminal voltage initially be \(V_i\). The currents supplied by P and Q satisfy \[ \frac{18-V_i}{2.0}+\frac{6.0-V_i}{2.0}=\frac{V_i}{3.0} \] Multiplying by six, \[ 3(18-V_i)+3(6.0-V_i)=2V_i \] \[ 72-6V_i=2V_i \] \[ V_i=9.0\,\text{V} \] \(\textbf{Determine Q's initial charging current:}\) \[ I_{Q,i}=\frac{6.0-9.0}{2.0} \] \[ I_{Q,i}=-1.50\,\text{A} \] Thus, Q initially charges at \(1.50\,\text{A}\). \(\textbf{Determine the new terminal voltage after Q's resistance changes:}\) The new internal resistance of Q is \(6.0\,\Omega\). Let the final terminal voltage be \(V_f\). Kirchhoff's junction rule becomes \[ \frac{18-V_f}{2.0}+\frac{6.0-V_f}{6.0}=\frac{V_f}{3.0} \] Multiplying by six, \[ 3(18-V_f)+(6.0-V_f)=2V_f \] \[ 54-3V_f+6.0-V_f=2V_f \] \[ 60=6V_f \] \[ V_f=10\,\text{V} \] \(\textbf{Determine the new charging current in Q:}\) \[ I_{Q,f}=\frac{6.0-10}{6.0} \] \[ I_{Q,f}=-\frac23\,\text{A} \] Q remains charging, but its charging-current magnitude decreases. \(\textbf{Compare external-load powers:}\) The external resistance is unchanged. Therefore, \[ \frac{P_{\text{load},f}}{P_{\text{load},i}}=\left(\frac{V_f}{V_i}\right)^2 \] \[ \frac{P_{\text{load},f}}{P_{\text{load},i}}=\left(\frac{10}{9.0}\right)^2 \] \[ \frac{P_{\text{load},f}}{P_{\text{load},i}}=\frac{100}{81} \] \(\textbf{Compare Q's chemical-energy storage rates:}\) Q's EMF remains constant. Its chemical-energy storage rate is proportional to the charging-current magnitude. Thus, \[ \frac{P_{\text{ch},f}}{P_{\text{ch},i}}=\frac{|I_{Q,f}|}{|I_{Q,i}|} \] \[ \frac{P_{\text{ch},f}}{P_{\text{ch},i}}=\frac{2/3}{3/2} \] \[ \frac{P_{\text{ch},f}}{P_{\text{ch},i}}=\frac49 \] \(\textbf{Physical interpretation:}\) The weaker cell initially receives current from the stronger cell. Increasing its internal resistance makes that charging path less effective. The reduced charging-current demand allows the common terminal voltage to rise. Consequently, the external resistor receives greater power even though Q stores chemical energy more slowly. An increase in one cell's internal resistance need not decrease the external-load power when that cell was initially being charged by another source. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_{\text{load},f}}{P_{\text{load},i}}=\frac{100}{81},\qquad\frac{P_{\text{ch},f}}{P_{\text{ch},i}}=\frac49} \]

Explanation: \(\textbf{Assign a common terminal-voltage reference:}\) Let \[ V=V_U-V_L \] The algebraic supplying current from P is \[ I_P=\frac{12-V}{2.0} \] After Q is reversed, its EMF is negative relative to the U-positive reference. Therefore, \[ I_Q=\frac{-6.0-V}{1.0} \] The external load carries \[ I_R=\frac V{3.0} \] \(\textbf{Apply Kirchhoff's junction rule:}\) \[ I_P+I_Q=I_R \] Substituting, \[ \frac{12-V}{2.0}+(-6.0-V)=\frac V{3.0} \] Multiplying by six, \[ 3(12-V)+6(-6.0-V)=2V \] \[ 36-3V-36-6V=2V \] \[ 11V=0 \] \[ V=0\,\text{V} \] \(\textbf{Determine external-load current and power:}\) \[ I_R=\frac V{3.0} \] \[ I_R=0\,\text{A} \] Therefore, \[ P_R=I_R^2R \] \[ P_R=0\,\text{W} \] \(\textbf{Determine the circulating cell currents:}\) For P, \[ I_P=\frac{12-0}{2.0} \] \[ I_P=6.0\,\text{A} \] For reversed Q, \[ I_Q=\frac{-6.0-0}{1.0} \] \[ I_Q=-6.0\,\text{A} \] The equal and opposite algebraic currents sum to zero at the external terminals. However, a circulating current of magnitude \(6.0\,\text{A}\) passes through the two cell branches. \(\textbf{Calculate internal heating in P:}\) \[ P_{\text{int},P}=I_P^2r_P \] \[ P_{\text{int},P}=(6.0)^2(2.0) \] \[ P_{\text{int},P}=72\,\text{W} \] \(\textbf{Calculate internal heating in Q:}\) \[ P_{\text{int},Q}=I_Q^2r_Q \] \[ P_{\text{int},Q}=(6.0)^2(1.0) \] \[ P_{\text{int},Q}=36\,\text{W} \] \(\textbf{Combine the internal losses:}\) \[ P_{\text{int,total}}=72+36 \] \[ P_{\text{int,total}}=108\,\text{W} \] \(\textbf{Verify the chemical-energy balance:}\) In the resulting circulating-current direction, both cells supply electrical energy through their EMFs. Their combined chemical-energy conversion rate is \[ P_{\text{chemical}}=(12)(6.0)+(6.0)(6.0) \] \[ P_{\text{chemical}}=108\,\text{W} \] This energy is entirely dissipated internally. The external resistor receives no current or electrical power. \(\textbf{Physical interpretation:}\) The two cells impose opposing terminal-voltage contributions after reversal. Their internal resistances permit a finite circulating current despite zero external terminal voltage. The external load therefore carries no current, but the cell assembly is not inactive. Zero external power does not imply zero chemical-energy conversion or zero internal heating. \(\textbf{Final answer:}\) \[ \boxed{P_R=0\,\text{W},\qquad P_{\text{int,total}}=108\,\text{W}} \]

Explanation: \(\textbf{Determine the complete source-branch parameters:}\) Branch P contains two aiding cells: \[ \mathcal E_P=6.0+6.0 \] \[ \mathcal E_P=12\,\text{V} \] Its internal resistance is \[ r_P=1.0+1.0 \] \[ r_P=2.0\,\Omega \] Branch Q contains two aiding EMFs and one opposing EMF: \[ \mathcal E_Q=4.0+4.0-4.0 \] \[ \mathcal E_Q=4.0\,\text{V} \] Its three internal resistances are in series: \[ r_Q=3.0\,\Omega \] \(\textbf{Express the two branch currents:}\) Let the common terminal potential difference be V. Taking currents supplied upwards into U as positive, \[ I_P=\frac{12-V}{2.0} \] \[ I_Q=\frac{4.0-V}{3.0} \] The external load draws \[ I_L=\frac V{4.0} \] \(\textbf{Apply the junction rule:}\) \[ I_P+I_Q=I_L \] Therefore, \[ \frac{12-V}{2.0}+\frac{4.0-V}{3.0}=\frac V{4.0} \] Multiplying by twelve, \[ 6(12-V)+4(4.0-V)=3V \] \[ 72-6V+16-4V=3V \] \[ 88=13V \] \[ V=\frac{88}{13}\,\text{V} \] \(\textbf{Calculate the external-load current:}\) \[ I_L=\frac{V}{4.0} \] \[ I_L=\frac{88/13}{4.0} \] \[ I_L=\frac{22}{13}\,\text{A} \] \(\textbf{Determine the current direction through branch Q:}\) \[ I_Q=\frac{4.0-88/13}{3.0} \] \[ I_Q=\frac{52/13-88/13}{3.0} \] \[ I_Q=-\frac{12}{13}\,\text{A} \] The negative sign indicates that actual current flows downwards from U towards L through branch Q. \(\textbf{Interpret individual cell operation inside Q:}\) Two cells in Q have their positive terminals oriented towards U. The downward branch current enters their positive terminals, so those cells charge. The third cell has the opposite polarity. The same downward current passes through that reversed cell from its negative side towards its positive side. It therefore discharges and converts chemical energy into electrical energy. \(\textbf{Calculate the reversed cell's chemical-energy conversion rate:}\) The current magnitude through every series cell in Q is \[ I_{\text{branch},Q}=\frac{12}{13}\,\text{A} \] The reversed cell has EMF \(4.0\,\text{V}\). Thus, \[ P_{\text{reversed}}=\mathcal EI_{\text{branch},Q} \] \[ P_{\text{reversed}}=(4.0)\left(\frac{12}{13}\right) \] \[ P_{\text{reversed}}=\frac{48}{13}\,\text{W} \] \(\textbf{Physical interpretation:}\) The three cells in branch Q carry the same current because they are in series. However, identical current direction does not imply identical chemical-energy transfer direction. The two similarly oriented cells charge, while the reversed cell discharges. The branch as a whole receives electrical current from the upper junction, yet one of its individual cells supplies electrical energy. The operating state of each cell must be determined from its own polarity relative to the actual series current. \(\textbf{Final answer:}\) \[ \boxed{I_L=\frac{22}{13}\,\text{A},\qquad P_{\text{reversed}}=\frac{48}{13}\,\text{W}} \]

Explanation: \(\textbf{Express the current supplied by P:}\) For either operating condition, \[ I_P=\frac{\mathcal E_P-V}{r_P} \] The current through Q is known from its EMF, internal resistance and the measured common voltage. \(\textbf{Experiment I — Determine the load current:}\) \[ R_1=1.0\,\Omega,\qquad V_1=5.0\,\text{V} \] Therefore, \[ I_{L,1}=\frac{V_1}{R_1} \] \[ I_{L,1}=5.0\,\text{A} \] \(\textbf{Determine Q's current in experiment I:}\) \[ I_{Q,1}=\frac{6.0-5.0}{2.0} \] \[ I_{Q,1}=0.50\,\text{A} \] Q supplies current in this condition. \(\textbf{Infer the corresponding current from P:}\) \[ I_{P,1}=I_{L,1}-I_{Q,1} \] \[ I_{P,1}=5.0-0.50 \] \[ I_{P,1}=4.50\,\text{A} \] Hence, \[ \mathcal E_P=5.0+4.50r_P \] \(\textbf{Experiment II — Determine the changed load current:}\) \[ R_2=4.0\,\Omega,\qquad V_2=8.0\,\text{V} \] Thus, \[ I_{L,2}=\frac{8.0}{4.0} \] \[ I_{L,2}=2.0\,\text{A} \] \(\textbf{Determine Q's changed current direction:}\) \[ I_{Q,2}=\frac{6.0-8.0}{2.0} \] \[ I_{Q,2}=-1.0\,\text{A} \] Q is charging at \(1.0\,\text{A}\). \(\textbf{Determine the current supplied by P in experiment II:}\) \[ I_{P,2}+I_{Q,2}=I_{L,2} \] \[ I_{P,2}-1.0=2.0 \] \[ I_{P,2}=3.0\,\text{A} \] Therefore, \[ \mathcal E_P=8.0+3.0r_P \] \(\textbf{Combine the two terminal-voltage relations:}\) The EMF and internal resistance of P are unchanged. Thus, \[ 5.0+4.50r_P=8.0+3.0r_P \] \[ 1.50r_P=3.0 \] \[ r_P=2.0\,\Omega \] \(\textbf{Recover the EMF:}\) \[ \mathcal E_P=8.0+3.0(2.0) \] \[ \mathcal E_P=14\,\text{V} \] \(\textbf{Physical interpretation:}\) In the first experiment, both cells supply the external load. In the second experiment, the higher terminal voltage causes Q to charge. The signed current supplied by Q must therefore be included correctly in the junction equation. Two distinct operating conditions provide two independent relations for the unknown cell's EMF and internal resistance. The observed change in Q's operating direction is essential to obtaining the correct result. \(\textbf{Final answer:}\) \[ \boxed{r_P=2.0\,\Omega,\qquad\mathcal E_P=14\,\text{V}} \]

Explanation: Statement I is correct because the cells share one terminal potential difference. For each real cell, its branch current depends on the difference between its EMF and that common voltage. With no external load, the algebraic sum of the two branch currents must be zero. For unequal positive EMFs and finite positive internal resistances, the resulting common voltage is a positive conductance-weighted average of the two EMFs. Such an average lies strictly between the two unequal values. Statement II is incorrect because the parallel connections themselves provide a closed conducting path between the cells. The absence of an external load does not interrupt that circulating-current path. Statement III is correct because the common voltage is below the higher EMF but above the lower EMF. Current therefore leaves the positive terminal of the higher-EMF cell and enters the positive terminal of the lower-EMF cell. The first discharges, while the second charges. Statement IV is correct because the circulating-current magnitude is the EMF difference divided by the sum of the two internal resistances. Increasing the lower cell's internal resistance increases that total resistance and decreases the circulating current. The internal voltage drop in the higher-EMF cell consequently decreases. Its loaded terminal voltage rises towards its own EMF. The terminal voltage does not move towards the lower EMF, because that cell's influence on the common terminal voltage becomes weaker as its internal resistance increases. The circulating-current energy is divided between internal heating and chemical-energy storage in the lower-EMF cell. The correct combination is Statements I, III and IV.

Explanation: \(\textbf{Determine the initial common terminal voltage:}\) Let the common voltage before shorting be \(V_i\). The two cells supply algebraic currents \[ I_P=\frac{12-V_i}{2.0} \] and \[ I_Q=\frac{6.0-V_i}{2.0} \] The external resistor draws \[ I_R=\frac{V_i}{3.0} \] \(\textbf{Apply the original junction equation:}\) \[ \frac{12-V_i}{2.0}+\frac{6.0-V_i}{2.0}=\frac{V_i}{3.0} \] Multiplying by six, \[ 3(12-V_i)+3(6.0-V_i)=2V_i \] \[ 54-6V_i=2V_i \] \[ V_i=\frac{27}{4}\,\text{V} \] \(\textbf{Determine Q's original charging current:}\) \[ I_{Q,i}=\frac{6.0-27/4}{2.0} \] \[ I_{Q,i}=\frac{24/4-27/4}{2.0} \] \[ I_{Q,i}=-\frac38\,\text{A} \] Q initially receives charging current. \(\textbf{Calculate its initial chemical-energy storage rate:}\) \[ P_{Q,i}=\mathcal E_Q|I_{Q,i}| \] \[ P_{Q,i}=(6.0)\left(\frac38\right) \] \[ P_{Q,i}=\frac94\,\text{W} \] \(\textbf{Determine P's initial supplying current:}\) \[ I_{P,i}=\frac{12-27/4}{2.0} \] \[ I_{P,i}=\frac{21}{8}\,\text{A} \] \(\textbf{Apply the ideal short-circuit condition:}\) An ideal wire directly joining the common terminals imposes \[ V_f=0 \] The individual cell currents are now determined by their EMFs and internal resistances. \(\textbf{Determine both short-circuit currents:}\) For P, \[ I_{P,f}=\frac{12}{2.0} \] \[ I_{P,f}=6.0\,\text{A} \] For Q, \[ I_{Q,f}=\frac{6.0}{2.0} \] \[ I_{Q,f}=3.0\,\text{A} \] Both cells now discharge into the short-circuit path. \(\textbf{Determine Q's final chemical-energy conversion rate:}\) \[ P_{Q,f}=\mathcal E_QI_{Q,f} \] \[ P_{Q,f}=(6.0)(3.0) \] \[ P_{Q,f}=18\,\text{W} \] \(\textbf{Compare chemical-energy rates:}\) \[ \frac{P_{Q,f}}{P_{Q,i}}=\frac{18}{9/4} \] \[ \frac{P_{Q,f}}{P_{Q,i}}=8 \] \(\textbf{Compare P's supplying currents:}\) \[ \frac{I_{P,f}}{I_{P,i}}=\frac{6.0}{21/8} \] \[ \frac{I_{P,f}}{I_{P,i}}=\frac{16}{7} \] \(\textbf{Physical interpretation:}\) Initially, the common voltage exceeds Q's EMF, causing Q to charge. The ideal short circuit reduces the common terminal voltage to zero. Both cell EMFs then exceed the terminal voltage, so both cells supply current. A short circuit can therefore reverse the operating state of a weaker cell from charging to discharging. The ideal wire dissipates no power, but substantial energy is dissipated in the cells' internal resistances. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_{Q,f}}{P_{Q,i}}=8,\qquad\frac{I_{P,f}}{I_{P,i}}=\frac{16}{7}} \]

Explanation: \(\textbf{Determine the initial circulating current:}\) The two unequal-EMF cells remain connected to each other even without an external load. The circulating-current magnitude is \[ I_c=\frac{\mathcal E_P-\mathcal E_Q}{r_P+r_Q} \] Substituting, \[ I_c=\frac{12-6.0}{2.0+2.0} \] \[ I_c=1.50\,\text{A} \] P discharges while Q charges. Therefore, \[ I_{\text{ch},i}=1.50\,\text{A} \] \(\textbf{Determine the initial common terminal voltage:}\) Using P's terminal relation, \[ V_i=\mathcal E_P-I_cr_P \] \[ V_i=12-(1.50)(2.0) \] \[ V_i=9.0\,\text{V} \] \(\textbf{Account for the voltmeter's finite resistance:}\) After connection, the voltmeter provides an external conducting path. Its current is \[ I_V=\frac V{3.0} \] The two cell currents are \[ I_P=\frac{12-V}{2.0} \] and \[ I_Q=\frac{6.0-V}{2.0} \] \(\textbf{Apply the junction rule with the voltmeter connected:}\) \[ I_P+I_Q=I_V \] Therefore, \[ \frac{12-V}{2.0}+\frac{6.0-V}{2.0}=\frac V{3.0} \] Multiplying by six, \[ 3(12-V)+3(6.0-V)=2V \] \[ 54-6V=2V \] \[ V=\frac{27}{4}\,\text{V} \] Thus, the voltmeter reads \[ V_{\text{meter}}=6.75\,\text{V} \] \(\textbf{Determine Q's changed current:}\) \[ I_{Q,f}=\frac{6.0-27/4}{2.0} \] \[ I_{Q,f}=-\frac38\,\text{A} \] The negative sign indicates that Q remains charging. Its charging-current magnitude is \[ I_{\text{ch},f}=\frac38\,\text{A} \] \(\textbf{Compare charging-current magnitudes:}\) \[ \frac{I_{\text{ch},f}}{I_{\text{ch},i}}=\frac{3/8}{3/2} \] \[ \frac{I_{\text{ch},f}}{I_{\text{ch},i}}=\frac14 \] \(\textbf{Physical interpretation:}\) The initial open-terminal voltage is determined by circulating current between the unequal cells. Connecting a finite-resistance voltmeter draws external current and changes the common terminal voltage. The resulting lower voltage reduces the current charging Q. The voltmeter reading is therefore not necessarily the same as the voltage that existed before the instrument was attached. An ideal voltmeter would draw no current and would not produce this loading effect. \(\textbf{Final answer:}\) \[ \boxed{V_{\text{meter}}=6.75\,\text{V},\qquad\frac{I_{\text{ch},f}}{I_{\text{ch},i}}=\frac14} \]

Explanation: \(\textbf{Determine the initial load voltage:}\) Initially, only P supplies the external resistor. The circuit current is \[ I_i=\frac{\mathcal E_P}{R+r_P} \] Substituting, \[ I_i=\frac{12}{4.0+2.0} \] \[ I_i=2.0\,\text{A} \] Therefore, \[ V_i=I_iR \] \[ V_i=(2.0)(4.0) \] \[ V_i=8.0\,\text{V} \] \(\textbf{Calculate the initial external power:}\) \[ P_i=\frac{V_i^2}{R} \] \[ P_i=\frac{(8.0)^2}{4.0} \] \[ P_i=16\,\text{W} \] \(\textbf{Represent the new parallel-cell arrangement:}\) After connecting Q, the algebraic cell currents are \[ I_P=\frac{12-V_f}{2.0} \] \[ I_Q=\frac{6.0-V_f}{2.0} \] The external resistor still draws \[ I_R=\frac{V_f}{4.0} \] \(\textbf{Apply the new junction equation:}\) \[ I_P+I_Q=I_R \] Substituting, \[ \frac{12-V_f}{2.0}+\frac{6.0-V_f}{2.0}=\frac{V_f}{4.0} \] Multiplying by four, \[ 2(12-V_f)+2(6.0-V_f)=V_f \] \[ 24-2V_f+12-2V_f=V_f \] \[ 36=5V_f \] \[ V_f=\frac{36}{5}\,\text{V} \] \(\textbf{Determine the final external-load power:}\) \[ P_f=\frac{V_f^2}{R} \] \[ P_f=\frac{(36/5)^2}{4.0} \] \[ P_f=\frac{324}{25}\,\text{W} \] \(\textbf{Compare the external powers:}\) \[ \frac{P_f}{P_i}=\frac{324/25}{16} \] \[ \frac{P_f}{P_i}=\frac{81}{100} \] \(\textbf{Determine Q's operating direction:}\) \[ I_Q=\frac{6.0-36/5}{2.0} \] \[ I_Q=\frac{30/5-36/5}{2.0} \] \[ I_Q=-\frac35\,\text{A} \] Thus, Q receives charging current of magnitude \(0.60\,\text{A}\). \(\textbf{Calculate its chemical-energy storage rate:}\) \[ P_{\text{chemical},Q}=\mathcal E_QI_{\text{ch},Q} \] \[ P_{\text{chemical},Q}=(6.0)\left(\frac35\right) \] \[ P_{\text{chemical},Q}=\frac{18}{5}\,\text{W} \] \[ P_{\text{chemical},Q}=3.6\,\text{W} \] \(\textbf{Physical interpretation:}\) Before Q is connected, the loaded terminal voltage of P exceeds Q's EMF. Connecting Q therefore creates a charging path rather than providing an additional discharging source. The current diverted into Q changes the operating voltage of P. The external resistor receives a lower terminal voltage and consequently dissipates less power. Adding a second cell in parallel can therefore decrease external-load power when the added cell has a sufficiently low EMF. The result depends on the actual EMFs, internal resistances and load, not simply on the number of connected cells. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_f}{P_i}=\frac{81}{100},\qquad P_{\text{chemical},Q}=3.6\,\text{W}} \]

Explanation: \(\textbf{Identify the original balanced arm ratios:}\) The original left upper resistance is \[ P_i=3.0+3.0 \] \[ P_i=6.0\,\Omega \] The left lower resistance is \(9.0\,\Omega\). The corresponding right-arm ratio is \[ \frac RS=\frac{4.0}{6.0}=\frac23 \] The original left ratio is also \[ \frac{P_i}{Q_i}=\frac{6.0}{9.0}=\frac23 \] Thus, the bridge is initially balanced. \(\textbf{Determine the modified upper left resistance:}\) One \(3.0\,\Omega\) resistor is now shunted by another \(3.0\,\Omega\) resistor. Their parallel equivalent is \[ R_p=\frac{(3.0)(3.0)}{3.0+3.0} \] \[ R_p=1.50\,\Omega \] The other original \(3.0\,\Omega\) resistor remains in series. Therefore, \[ P_f=3.0+1.50 \] \[ P_f=4.50\,\Omega \] \(\textbf{Find the required lower left resistance:}\) To restore balance, \[ \frac{P_f}{Q_f}=\frac RS \] Substituting, \[ \frac{4.50}{Q_f}=\frac23 \] \[ Q_f=6.75\,\Omega \] \(\textbf{Recover the compensating shunt resistance:}\) Let x be the resistance added across the original \(9.0\,\Omega\) arm. Then, \[ \frac{9.0x}{9.0+x}=6.75 \] \[ 9.0x=60.75+6.75x \] \[ 2.25x=60.75 \] \[ x=27\,\Omega \] \(\textbf{Determine the original bridge equivalent resistance:}\) At balance, the galvanometer carries no current. The two complete conducting paths initially have resistances \[ R_{L,i}=6.0+9.0=15\,\Omega \] \[ R_{R,i}=4.0+6.0=10\,\Omega \] Their parallel equivalent is \[ R_{\text{eq},i}=\frac{(15)(10)}{15+10} \] \[ R_{\text{eq},i}=6.0\,\Omega \] \(\textbf{Determine the restored equivalent resistance:}\) After both modifications, the left path has resistance \[ R_{L,f}=4.50+6.75 \] \[ R_{L,f}=11.25\,\Omega \] The right path remains \(10\,\Omega\). Therefore, \[ R_{\text{eq},f}=\frac{(11.25)(10)}{11.25+10} \] \[ R_{\text{eq},f}=\frac{90}{17}\,\Omega \] \(\textbf{Compare source currents:}\) The ideal source voltage remains constant. Hence, \[ \frac{I_f}{I_i}=\frac{R_{\text{eq},i}}{R_{\text{eq},f}} \] \[ \frac{I_f}{I_i}=\frac{6.0}{90/17} \] \[ \frac{I_f}{I_i}=\frac{17}{15} \] \(\textbf{Physical interpretation:}\) The first shunt reduces the upper left resistance and destroys the original balance. The second shunt restores the required ratio by reducing the lower left resistance as well. The galvanometer again carries zero current, but the total resistance of the bridge is smaller than initially. Consequently, the ideal source supplies a larger current. Restoring a bridge null does not necessarily restore the original source current. \(\textbf{Final answer:}\) \[ \boxed{x=27\,\Omega,\qquad\frac{I_f}{I_i}=\frac{17}{15}} \]

Explanation: \(\textbf{Determine the heated left-arm resistances:}\) For the stated linear temperature dependence, \[ R_T=R_0(1+\alpha\Delta T) \] The upper left resistor becomes \[ P_f=3.0[1+(4.0\times10^{-3})(100)] \] \[ P_f=4.20\,\Omega \] The lower left resistor becomes \[ Q_f=4.0[1+(2.0\times10^{-3})(100)] \] \[ Q_f=4.80\,\Omega \] \(\textbf{Determine the heated lower right resistance:}\) \[ S_f=8.0[1+(1.0\times10^{-3})(100)] \] \[ S_f=8.80\,\Omega \] \(\textbf{Use the preserved bridge balance:}\) At balance, \[ \frac{P_f}{Q_f}=\frac{R_f}{S_f} \] Therefore, \[ \frac{4.20}{4.80}=\frac{R_f}{8.80} \] \[ R_f=8.80\left(\frac{4.20}{4.80}\right) \] \[ R_f=7.70\,\Omega \] \(\textbf{Recover the unknown temperature coefficient:}\) The upper right resistor initially had resistance \(6.0\,\Omega\). Thus, \[ 7.70=6.0(1+100\alpha) \] \[ 1+100\alpha=\frac{77}{60} \] \[ 100\alpha=\frac{17}{60} \] \[ \alpha=\frac{17}{6000}\,\text{K}^{-1} \] \(\textbf{Compare the two conducting-path currents:}\) Let the unchanged ideal source voltage be V. The heated left path has resistance \[ R_L=4.20+4.80=9.0\,\Omega \] The heated right path has resistance \[ R_R=7.70+8.80=16.50\,\Omega \] At balance, the galvanometer current is zero. Therefore, \[ I_L=\frac V9 \] and \[ I_R=\frac V{16.50} \] \(\textbf{Compare heating in the two upper arms:}\) The upper left resistor dissipates \[ P_{\text{upper left}}=I_L^2(4.20) \] The upper right resistor dissipates \[ P_{\text{upper right}}=I_R^2(7.70) \] Therefore, \[ \frac{P_{\text{upper right}}}{P_{\text{upper left}}} =\left(\frac{I_R}{I_L}\right)^2\frac{7.70}{4.20} \] Substituting, \[ \frac{P_{\text{upper right}}}{P_{\text{upper left}}} =\left(\frac{9.0}{16.50}\right)^2\left(\frac{7.70}{4.20}\right) \] \[ =\left(\frac6{11}\right)^2\left(\frac{11}{6}\right) \] \[ =\frac6{11} \] \(\textbf{Physical interpretation:}\) The four resistors do not need equal temperature coefficients for the bridge to remain balanced. Their heated resistance ratios must instead remain equal. Although the upper right resistor has greater resistance than the upper left resistor, it carries a smaller current because its complete conducting path has greater resistance. The difference in branch currents determines the final heating comparison. \(\textbf{Final answer:}\) \[ \boxed{\alpha=\frac{17}{6000}\,\text{K}^{-1},\qquad \frac{P_{\text{upper right}}}{P_{\text{upper left}}}=\frac6{11}} \]

Explanation: \(\textbf{Verify the original bridge balance:}\) The four arm resistances satisfy \[ \frac{2.0}{4.0}=\frac{3.0}{6.0} \] Therefore, B and C are equipotential when the source is connected between A and D. The galvanometer initially carries zero current. \(\textbf{Examine the interchanged source and detector terminals:}\) In experiment II, the source is connected between B and C. The conducting path B–A–C has resistance \[ R_{\text{upper}}=2.0+3.0 \] \[ R_{\text{upper}}=5.0\,\Omega \] The other conducting path B–D–C has resistance \[ R_{\text{lower}}=4.0+6.0 \] \[ R_{\text{lower}}=10\,\Omega \] \(\textbf{Check the new detector-terminal potentials:}\) Take C as zero potential and B as V. With the A–D detector branch open, the potential at A is \[ V_A=\frac{3.0}{2.0+3.0}V \] \[ V_A=\frac35V \] Similarly, the potential at D is \[ V_D=\frac{6.0}{4.0+6.0}V \] \[ V_D=\frac35V \] Thus, \[ V_A=V_D \] Connecting the galvanometer between A and D therefore produces zero current. \(\textbf{Determine the original source equivalent resistance:}\) In experiment I, the A–B–D path has resistance \[ R_{L,i}=2.0+4.0=6.0\,\Omega \] The A–C–D path has resistance \[ R_{R,i}=3.0+6.0=9.0\,\Omega \] Their parallel equivalent is \[ R_{\text{eq},i}=\frac{(6.0)(9.0)}{6.0+9.0} \] \[ R_{\text{eq},i}=\frac{18}{5}\,\Omega \] \(\textbf{Determine the equivalent resistance after interchange:}\) The new conducting paths have resistances \(5.0\,\Omega\) and \(10\,\Omega\). Therefore, \[ R_{\text{eq},f}=\frac{(5.0)(10)}{5.0+10} \] \[ R_{\text{eq},f}=\frac{10}{3}\,\Omega \] \(\textbf{Compare the source currents:}\) The ideal source voltage magnitude is unchanged. Thus, \[ \frac{I_f}{I_i}=\frac{R_{\text{eq},i}}{R_{\text{eq},f}} \] \[ \frac{I_f}{I_i}=\frac{18/5}{10/3} \] \[ \frac{I_f}{I_i}=\frac{27}{25} \] \(\textbf{Physical interpretation:}\) The balanced bridge remains a null network when the supply and detector are exchanged between its two diagonals. However, the equivalent resistance seen by the source is different because the source now connects a different pair of junctions. The null condition is preserved while the source current increases. The detector reading and the input resistance are therefore distinct properties of the bridge. \(\textbf{Final answer:}\) \[ \boxed{I_{G,\mathrm{II}}=0,\qquad\frac{I_{\mathrm{II}}}{I_{\mathrm I}}=\frac{27}{25}} \]

Explanation: The direction of galvanometer current is determined by the relative potentials of the intermediate junctions. A useful first step is to compare their potentials with the detector branch open. Connecting a finite positive-resistance galvanometer changes the actual potentials, but not the direction established by the open-detector potential difference in this passive network. Originally, both intermediate junctions are at the same potential. In experiment I, the resistance above B increases while the resistance below B is unchanged. The left divider therefore assigns a smaller fraction of the source voltage to B. The right divider remains unchanged, so C stays at its original potential. Consequently, C is at higher potential than B, and current flows from C to B. In experiment II, the resistance below C decreases while the upper right resistance remains unchanged. The potential of C consequently decreases. The left divider remains unchanged, so B stays at its original potential. Thus, B is at higher potential than C, and current flows from B to C. In experiment III, both modifications act to lower an intermediate-junction potential, so their relative effects must be compared. With A–B increased to \(3.0\,\Omega\), the open-detector potential at B is \[ V_B=12\left(\frac{4.0}{3.0+4.0}\right)=\frac{48}{7}\,\text{V} \] With C–D decreased to \(3.0\,\Omega\), the open-detector potential at C is \[ V_C=12\left(\frac{3.0}{3.0+3.0}\right)=6.0\,\text{V} \] Since \(48/7\) exceeds \(6.0\), B is at higher potential than C. The galvanometer current in experiment III therefore flows from B to C. This comparison shows why the direction under two simultaneous changes cannot be inferred merely by noting that both junction potentials decrease. The correct sequence is C to B, B to C and B to C.

Explanation: \(\textbf{Use the measured galvanometer current:}\) Choose \[ V_A=12\,\text{V},\qquad V_D=0 \] Let \[ V_B=x,\qquad V_C=y \] The galvanometer current flows from B to C. Therefore, \[ \frac{x-y}{4.0}=0.50 \] \[ x-y=2.0\,\text{V} \] \(\textbf{Apply the junction rule at B:}\) The current arriving from A must equal the currents leaving towards D and C: \[ \frac{12-x}{2.0}=\frac x{4.0}+0.50 \] Multiplying by four, \[ 2(12-x)=x+2.0 \] \[ 24-2x=x+2.0 \] \[ 3x=22 \] \[ x=\frac{22}{3}\,\text{V} \] \(\textbf{Determine the potential at C:}\) Using the galvanometer voltage relation, \[ y=x-2.0 \] \[ y=\frac{22}{3}-2.0 \] \[ y=\frac{16}{3}\,\text{V} \] \(\textbf{Apply the junction rule at C:}\) Current arriving from A through X and from B through the galvanometer must equal the current leaving C towards D: \[ \frac{12-y}{X}+0.50=\frac y{4.0} \] Therefore, \[ \frac{12-y}{X}=\frac y{4.0}-0.50 \] Substituting, \[ \frac{12-16/3}{X}=\frac{16/3}{4.0}-0.50 \] \[ \frac{20/3}{X}=\frac43-\frac12 \] \[ \frac{20/3}{X}=\frac56 \] Hence, \[ X=\frac{20/3}{5/6} \] \[ X=8.0\,\Omega \] \(\textbf{Determine current through the left upper arm:}\) \[ I_{AB}=\frac{12-x}{2.0} \] \[ I_{AB}=\frac{12-22/3}{2.0} \] \[ I_{AB}=\frac73\,\text{A} \] \(\textbf{Determine current through the right upper arm:}\) \[ I_{AC}=\frac{12-y}{X} \] \[ I_{AC}=\frac{12-16/3}{8.0} \] \[ I_{AC}=\frac56\,\text{A} \] \(\textbf{Apply the source-terminal junction rule:}\) \[ I_{\text{source}}=I_{AB}+I_{AC} \] \[ I_{\text{source}}=\frac73+\frac56 \] \[ I_{\text{source}}=\frac{19}{6}\,\text{A} \] \(\textbf{Physical interpretation:}\) The finite galvanometer current changes the potentials at both intermediate junctions. It cannot be treated as though the detector branch were absent. The measured galvanometer current supplies a potential-difference relation, which combines with two junction equations to determine the unknown resistance. The resulting individual upper-arm currents then determine the source current. \(\textbf{Final answer:}\) \[ \boxed{X=8.0\,\Omega,\qquad I_{\text{source}}=\frac{19}{6}\,\text{A}} \]

Explanation: Initially, the corresponding upper-to-lower arm ratios are equal. Therefore, the two galvanometer terminals are initially equipotential. Adding the same resistance in parallel to different original resistances does not multiply those original resistances by a common factor. Each new arm resistance must be calculated using the parallel-combination relation. The modified A–B resistance is \[ P'=\frac{(2.0)(2.0)}{2.0+2.0}=1.0\,\Omega \] The modified B–D resistance is \[ Q'=\frac{(4.0)(2.0)}{4.0+2.0}=\frac43\,\Omega \] The modified A–C resistance is \[ R'=\frac{(3.0)(2.0)}{3.0+2.0}=\frac65\,\Omega \] The modified C–D resistance is \[ S'=\frac{(6.0)(2.0)}{6.0+2.0}=\frac32\,\Omega \] The new left-arm ratio is \[ \frac{P'}{Q'}=\frac34 \] The new right-arm ratio is \[ \frac{R'}{S'}=\frac45 \] These ratios differ, so the bridge is no longer balanced. To determine current direction, compare the intermediate potentials with the detector temporarily open. Relative to the full supply voltage V, \[ V_B=\frac{Q'}{P'+Q'}V=\frac47V \] Similarly, \[ V_C=\frac{S'}{R'+S'}V=\frac59V \] Since \(4/7\) exceeds \(5/9\), B is initially at higher potential than C. Connecting the finite-resistance galvanometer therefore produces current from B towards C. The misconception is that equal numerical shunts produce proportional changes in unequal resistances. They do not. The bridge becomes unbalanced, and its galvanometer current flows from B to C.