Current Electricity MCQs with Answers – Part 6 (NEET Physics)

Explanation: The two cells share a common terminal potential difference, but their currents are not necessarily equal. \(\textbf{Determine the common terminal voltage:}\) Let that terminal voltage be \(V\). The algebraic supplying currents from P and Q are \[ I_P=\frac{12-V}{2},\qquad I_Q=\frac{6-V}{2} \] The external resistor carries current \[ I_{\text{load}}=\frac V8 \] Kirchhoff's junction rule requires the sum of the cell currents to equal the load current. \[ \frac{12-V}{2}+\frac{6-V}{2}=\frac V8 \] Multiplying through and solving, \[ 72-8V=V \] \[ V=8.0\,\text{V} \] \(\textbf{Determine which cell supplies or receives current:}\) For cell P, \[ I_P=\frac{12-8.0}{2.0}=2.0\,\text{A} \] For cell Q, \[ I_Q=\frac{6.0-8.0}{2.0}=-1.0\,\text{A} \] The negative algebraic supplying current means Q actually receives \(1.0\,\text{A}\) through its positive terminal. The external resistor carries \[ I_{\text{load}}=\frac{8.0}{8.0}=1.0\,\text{A} \] \(\textbf{Check the power statements:}\) Statement I is correct because Q receives terminal electrical power \[ P_{Q,\text{terminal}}=(8.0)(1.0)=8.0\,\text{W} \] Statement II is correct because P generates electrical power at the rate \[ P_{P,\text{emf}}=(12)(2.0)=24\,\text{W} \] Its internal resistance dissipates \[ P_{P,\text{internal}}=(2.0)^2(2.0)=8.0\,\text{W} \] This leaves \(16\,\text{W}\) delivered through P's terminals. Statement III is correct because the external resistor carries \(1.0\,\text{A}\) and dissipates \[ P_{\text{load}}=(1.0)^2(8.0)=8.0\,\text{W} \] The charging cell Q stores chemical energy at the rate \[ P_{Q,\text{chemical}}=(6.0)(1.0)=6.0\,\text{W} \] The remaining \(2.0\,\text{W}\) of its terminal input is dissipated in its internal resistance. Statement IV is incorrect because equal terminal voltage does not imply equal currents when the cell EMFs differ. The signed currents must instead satisfy their individual voltage-current relations and the common junction equation. The complete power balance is \[ 24=8+8+6+2\quad(\text{W}) \] This confirms that Statements I, II and III are correct.

Explanation: \(\textbf{Express the branch-current relations:}\) Let \[ V=V_U-V_L \] The algebraic currents supplied upwards by the real cells are \[ I_P=\frac{12-V}{2.0} \] \[ I_Q=\frac{6.0-V}{4.0} \] The external resistor carries \[ I_R=\frac V{4.0} \] \(\textbf{Experiment A — Current source disconnected:}\) Only the two real cells supply algebraic current at U. Kirchhoff's junction rule gives \[ I_P+I_Q=I_R \] Therefore, \[ \frac{12-V_A}{2.0}+\frac{6.0-V_A}{4.0}=\frac{V_A}{4.0} \] Multiplying by four, \[ 2(12-V_A)+(6.0-V_A)=V_A \] \[ 30-3V_A=V_A \] \[ V_A=7.5\,\text{V} \] \(\textbf{Determine Q's charging current in experiment A:}\) \[ I_{Q,A}=\frac{6.0-7.5}{4.0} \] \[ I_{Q,A}=-\frac38\,\text{A} \] Thus, Q receives charging current of magnitude \(3/8\,\text{A}\). \(\textbf{Experiment B — Include the injected current:}\) The current source now delivers \(2.0\,\text{A}\) into U. The junction equation becomes \[ I_P+I_Q+2.0=I_R \] Substituting, \[ \frac{12-V_B}{2.0}+\frac{6.0-V_B}{4.0}+2.0=\frac{V_B}{4.0} \] Multiplying by four, \[ 2(12-V_B)+(6.0-V_B)+8.0=V_B \] \[ 38-3V_B=V_B \] \[ V_B=9.5\,\text{V} \] \(\textbf{Determine Q's charging current in experiment B:}\) \[ I_{Q,B}=\frac{6.0-9.5}{4.0} \] \[ I_{Q,B}=-\frac78\,\text{A} \] Q continues charging, but its charging-current magnitude increases. \(\textbf{Compare chemical-energy storage rates:}\) For unchanged EMF, \[ P_{\text{chemical}}=\mathcal E_QI_{\text{ch}} \] Therefore, \[ \frac{P_{\text{chemical},B}}{P_{\text{chemical},A}} =\frac{I_{\text{ch},B}}{I_{\text{ch},A}} \] \[ \frac{P_{\text{chemical},B}}{P_{\text{chemical},A}} =\frac{7/8}{3/8} \] \[ \frac{P_{\text{chemical},B}}{P_{\text{chemical},A}}=\frac73 \] \(\textbf{Verify experiment B's junction balance:}\) Cell P supplies \[ I_{P,B}=\frac{12-9.5}{2.0} \] \[ I_{P,B}=1.25\,\text{A} \] The load draws \[ I_{R,B}=\frac{9.5}{4.0} \] \[ I_{R,B}=2.375\,\text{A} \] At U, \[ 1.25-0.875+2.0=2.375 \] Current conservation is satisfied. Injecting additional current raises the common terminal voltage and increases the charging current entering Q. The comparison requires solving the complete junction equation in both operating conditions rather than assuming that the injected current divides equally among the branches. \(\textbf{Final answer:}\) \[ \boxed{V_B=9.5\,\text{V},\quad\frac{P_{\text{chemical},B}}{P_{\text{chemical},A}}=\frac73} \]

Explanation: \(\textbf{Determine the initial equivalent conductance:}\) For resistors in parallel, conductances add: \[ G_i=\frac1{2.0}+\frac1{3.0}+\frac1{6.0} \] \[ G_i=\frac36+\frac26+\frac16 \] \[ G_i=1.0\,\text{S} \] The initial equivalent resistance is \[ R_i=\frac1{G_i} \] \[ R_i=1.0\,\Omega \] \(\textbf{Verify the initial voltage:}\) \[ V_i=I_{\text{source}}R_i \] \[ V_i=(6.0)(1.0) \] \[ V_i=6.0\,\text{V} \] This agrees with the original measurement. \(\textbf{Infer the final equivalent resistance:}\) The current source still supplies \(6.0\,\text{A}\). The new voltage is \(9.0\,\text{V}\). Therefore, \[ R_f=\frac{V_f}{I_{\text{source}}} \] \[ R_f=\frac{9.0}{6.0} \] \[ R_f=1.50\,\Omega \] \(\textbf{Determine the remaining conductance:}\) \[ G_f=\frac1{R_f} \] \[ G_f=\frac23\,\text{S} \] The conductance lost when the branch opened is \[ G_{\text{lost}}=G_i-G_f \] \[ G_{\text{lost}}=1.0-\frac23 \] \[ G_{\text{lost}}=\frac13\,\text{S} \] \(\textbf{Identify the faulty branch:}\) A resistance corresponding to this conductance is \[ R_{\text{open}}=\frac1{G_{\text{lost}}} \] \[ R_{\text{open}}=3.0\,\Omega \] Therefore, the \(3.0\,\Omega\) branch is open-circuited. \(\textbf{Compare power in the unchanged }2.0\,\Omega\textbf{ resistor:}\) This resistor remains connected across the source terminals. Its resistance does not change. Its power is \[ P_2=\frac{V^2}{2.0} \] Thus, \[ \frac{P_{2,f}}{P_{2,i}} =\left(\frac{V_f}{V_i}\right)^2 \] \[ \frac{P_{2,f}}{P_{2,i}} =\left(\frac{9.0}{6.0}\right)^2 \] \[ \frac{P_{2,f}}{P_{2,i}}=\frac94 \] \(\textbf{Check the final junction currents:}\) The remaining \(2.0\,\Omega\) branch carries \[ I_2=\frac{9.0}{2.0}=4.50\,\text{A} \] The remaining \(6.0\,\Omega\) branch carries \[ I_6=\frac{9.0}{6.0}=1.50\,\text{A} \] Their sum is \[ 4.50+1.50=6.0\,\text{A} \] This agrees with the imposed source current. Opening a parallel branch increases the equivalent resistance. Under a constant-current source, the terminal voltage consequently increases, raising the power dissipated in the surviving branches. The measured voltage change therefore both identifies the failed branch and predicts the redistributed heating. \(\textbf{Final answer:}\) \[ \boxed{R_{\text{open}}=3.0\,\Omega,\quad\frac{P_{2,f}}{P_{2,i}}=\frac94} \]

Explanation: \(\textbf{Use the initial opposing-source loop:}\) Initially, the net driving EMF is the difference between the cell EMFs: \[ \mathcal E_i=15-6.0 \] \[ \mathcal E_i=9.0\,\text{V} \] The external resistance and internal resistance of Q are in series. Therefore, \[ I_i=\frac{9.0}{3.0+r} \] \(\textbf{Recover Q's internal resistance:}\) The initial current magnitude is \(1.50\,\text{A}\). Thus, \[ 1.50=\frac{9.0}{3.0+r} \] \[ 3.0+r=\frac{9.0}{1.50} \] \[ 3.0+r=6.0 \] \[ r=3.0\,\Omega \] \(\textbf{Determine the initial external resistor power:}\) \[ P_i=I_i^2R \] \[ P_i=(1.50)^2(3.0) \] \[ P_i=6.75\,\text{W} \] \(\textbf{Analyse the source reversal:}\) After Q is reversed, its EMF aids P. The resultant EMF becomes \[ \mathcal E_f=15+6.0 \] \[ \mathcal E_f=21\,\text{V} \] Reversing a cell changes the sign of its EMF contribution but does not change its positive internal resistance. The complete loop resistance remains \[ R_{\text{complete}}=3.0+3.0 \] \[ R_{\text{complete}}=6.0\,\Omega \] \(\textbf{Determine the new loop current:}\) \[ I_f=\frac{21}{6.0} \] \[ I_f=3.50\,\text{A} \] \(\textbf{Calculate the new external power:}\) \[ P_f=I_f^2R \] \[ P_f=(3.50)^2(3.0) \] \[ P_f=36.75\,\text{W} \] \(\textbf{Compare external heating rates:}\) Because the external resistance is unchanged, \[ \frac{P_f}{P_i}=\left(\frac{I_f}{I_i}\right)^2 \] Substituting, \[ \frac{P_f}{P_i}=\left(\frac{3.50}{1.50}\right)^2 \] \[ \frac{P_f}{P_i}=\left(\frac73\right)^2 \] \[ \frac{P_f}{P_i}=\frac{49}{9} \] \(\textbf{Interpret the change in cell operation:}\) Initially, current enters Q's positive terminal and Q stores chemical energy. After reversal, the same two EMFs assist one another around the loop. Both cells then supply electrical energy. The reversal increases the net driving EMF while keeping total resistance unchanged, causing a greater current and a much larger external heating rate. The power ratio contains the square of the current ratio, so it cannot be obtained simply by dividing the two resultant EMFs. \(\textbf{Final answer:}\) \[ \boxed{r=3.0\,\Omega,\quad\frac{P_f}{P_i}=\frac{49}{9}} \]

Explanation: \(\textbf{Construct the equivalent source using Kirchhoff's laws:}\) Let the common terminal voltage be V. The algebraic currents supplied by the cells are \[ I_P=\frac{12-V}{2.0} \] \[ I_Q=\frac{6.0-V}{2.0} \] The external current is \[ I_L=I_P+I_Q \] Therefore, \[ I_L=\frac{18-2V}{2.0} \] \[ I_L=9.0-V \] Rearranging, \[ V=9.0-I_L(1.0) \] The parallel battery is therefore equivalent at its external terminals to a source having \[ \mathcal E_{\text{eq}}=9.0\,\text{V} \] and \[ r_{\text{eq}}=1.0\,\Omega \] \(\textbf{Express the external load power:}\) For positive R, \[ I_L=\frac{9.0}{R+1.0} \] The power dissipated in R is \[ P_R=I_L^2R \] \[ P_R=\frac{81R}{(R+1)^2} \] \(\textbf{Determine the maximum-power condition:}\) For positive R, \[ (R-1)^2\geq0 \] Therefore, \[ (R+1)^2\geq4R \] Consequently, \[ P_R=\frac{81R}{(R+1)^2}\leq\frac{81}{4} \] Equality occurs when \[ R=1.0\,\Omega \] This is the required external resistance. \(\textbf{Determine the load current at maximum power:}\) \[ I_L=\frac{9.0}{1.0+1.0} \] \[ I_L=4.50\,\text{A} \] \(\textbf{Determine the common terminal voltage:}\) \[ V=I_LR \] \[ V=(4.50)(1.0) \] \[ V=4.50\,\text{V} \] \(\textbf{Determine the current through Q:}\) \[ I_Q=\frac{\mathcal E_Q-V}{r_Q} \] \[ I_Q=\frac{6.0-4.50}{2.0} \] \[ I_Q=0.75\,\text{A} \] The positive sign indicates current leaving Q's positive terminal. Thus, Q is discharging. \(\textbf{Verify the individual cell currents:}\) Cell P supplies \[ I_P=\frac{12-4.50}{2.0} \] \[ I_P=3.75\,\text{A} \] The sum is \[ 3.75+0.75=4.50\,\text{A} \] This agrees with the load current. Although the lower-EMF cell can receive charging current under sufficiently light loading, it supplies current at the maximum-load-power setting. The loading condition determines the common terminal voltage and therefore the direction of current in each cell. \(\textbf{Final answer:}\) \[ \boxed{R=1.0\,\Omega,\quad I_Q=0.75\,\text{A}\text{ discharging}} \]

Explanation: The Assertion is true under the stated circuit conditions. An open switch interrupts the conducting path through its branch. Consequently, the steady current through the series resistor in that branch is zero. For a positive ohmic resistor, the potential drop is the product of current and resistance. Its potential drop therefore becomes zero when its current becomes zero. Its electrical heating rate is also zero because resistive power depends on the square of current. However, the branch remains connected to two different points of the surrounding powered network. Those points can be maintained at different potentials by sources and conducting paths elsewhere in the circuit. Since there is no potential drop across the series resistor, the potential difference between the branch terminals can appear across the open switch. An ideal voltmeter connected across the switch measures that potential difference without establishing a conducting current path. Kirchhoff's loop rule remains satisfied because the potential rises and drops along the complete closed paths of the powered network remain consistent. The open branch is not itself a closed conducting loop. The Reason correctly explains why zero branch current and nonzero switch voltage can occur simultaneously. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Initial circuit — Assign the unknown junction potentials:}\) The source fixes \[ V_A=12\,\text{V},\qquad V_D=0 \] Let the initial potentials be \[ V_{B,i}=x,\qquad V_{C,i}=y \] \(\textbf{Apply the initial junction rule at B:}\) Current entering B from A equals current leaving towards D and C: \[ \frac{12-x}{3}=\frac{x}{6}+\frac{x-y}{2} \] Multiplying by six, \[ 2(12-x)=x+3(x-y) \] \[ 24-2x=4x-3y \] Therefore, \[ 2x-y=8 \] \(\textbf{Apply the initial junction rule at C:}\) \[ \frac{12-y}{6}+\frac{x-y}{2}=\frac y3 \] Multiplying by six, \[ 12-y+3(x-y)=2y \] \[ 12+3x=6y \] Thus, \[ 2y=x+4 \] \(\textbf{Determine the original junction potentials:}\) From the first equation, \[ y=2x-8 \] Substituting into the second, \[ 2(2x-8)=x+4 \] \[ 3x=20 \] \[ x=\frac{20}{3}\,\text{V} \] Therefore, \[ y=2\left(\frac{20}{3}\right)-8 \] \[ y=\frac{16}{3}\,\text{V} \] \(\textbf{Calculate the initial connecting-branch current:}\) \[ I_{BC,i}=\frac{x-y}{2.0} \] \[ I_{BC,i}=\frac{20/3-16/3}{2.0} \] \[ I_{BC,i}=\frac23\,\text{A} \] The initial current flows from B to C. \(\textbf{Modified circuit — Account for the changed B–D resistor:}\) Only the resistance between B and D changes from \(6.0\,\Omega\) to \(1.0\,\Omega\). Let the final junction potentials be \[ V_{B,f}=u,\qquad V_{C,f}=v \] The new junction equation at B is \[ \frac{12-u}{3}=\frac{u}{1.0}+\frac{u-v}{2} \] Multiplying by six, \[ 2(12-u)=6u+3(u-v) \] \[ 24-2u=9u-3v \] Therefore, \[ 11u-3v=24 \] \(\textbf{Use the unchanged junction equation at C:}\) The resistors incident at C remain unchanged. Therefore, \[ \frac{12-v}{6}+\frac{u-v}{2}=\frac v3 \] Simplifying, \[ 2v=u+4 \] \(\textbf{Solve the modified network:}\) \[ v=\frac{u+4}{2} \] Substituting, \[ 11u-3\left(\frac{u+4}{2}\right)=24 \] Multiplying by two, \[ 22u-3u-12=48 \] \[ 19u=60 \] \[ u=\frac{60}{19}\,\text{V} \] Then, \[ v=\frac{60/19+4}{2} \] \[ v=\frac{68}{19}\,\text{V} \] \(\textbf{Determine the final B–C current:}\) \[ I_{BC,f}=\frac{u-v}{2.0} \] \[ I_{BC,f}=\frac{60/19-68/19}{2.0} \] \[ I_{BC,f}=-\frac4{19}\,\text{A} \] The negative sign means the final current flows from C to B. \(\textbf{Compare current magnitudes:}\) \[ \frac{|I_{BC,f}|}{|I_{BC,i}|} =\frac{4/19}{2/3} \] \[ \frac{|I_{BC,f}|}{|I_{BC,i}|}=\frac6{19} \] \(\textbf{Physical interpretation:}\) Decreasing the B–D resistance provides a stronger conducting path from B towards the reference junction. This lowers the potential of B relative to C sufficiently to reverse the current through their connecting resistor. The source polarity and the B–C resistor itself remain unchanged. A current can therefore reverse solely because another branch resistance changes. The two independent junction equations are necessary because B and C influence one another through the connecting resistor. \(\textbf{Final answer:}\) \[ \boxed{I_{BC,f}=\frac4{19}\,\text{A}\text{ from C to B},\quad\frac{|I_{BC,f}|}{|I_{BC,i}|}=\frac6{19}} \]

Explanation: \(\textbf{Relate the shared current to the mesh currents:}\) With downward current defined positive, \[ I_X=i_L-i_R \] The two outer resistances are each \(2.0\,\Omega\). \(\textbf{Write the general left mesh equation:}\) The left source contributes a \(12\,\text{V}\) rise. The left outer resistor contributes a drop \(2i_L\). The shared resistor contributes a drop \(XI_X\). Thus, \[ 12=2i_L+XI_X \] Rearranging, \[ i_L=\frac{12-XI_X}{2} \] \(\textbf{Write the general right mesh equation:}\) The right mesh traverses X upwards, opposite to the defined positive shared-current direction. Therefore, \[ E=2i_R-XI_X \] Hence, \[ i_R=\frac{E+XI_X}{2} \] \(\textbf{Combine the mesh-current relations:}\) Using \(I_X=i_L-i_R\), \[ I_X=\frac{12-XI_X}{2}-\frac{E+XI_X}{2} \] Multiplying by two, \[ 2I_X=12-XI_X-E-XI_X \] \[ (2+2X)I_X=12-E \] Therefore, \[ I_X=\frac{12-E}{2+2X} \] \(\textbf{Initial shared-branch current:}\) Initially, \[ E_i=8.0\,\text{V},\qquad X_i=2.0\,\Omega \] Substituting, \[ I_{X,i}=\frac{12-8.0}{2+2(2.0)} \] \[ I_{X,i}=\frac4{6.0} \] \[ I_{X,i}=+\frac23\,\text{A} \] The initial current flows downwards. \(\textbf{Apply the reversal and equal-magnitude requirement:}\) The new shared current must be \[ I_{X,f}=-\frac23\,\text{A} \] The right source now has \[ E_f=14\,\text{V} \] Therefore, \[ -\frac23=\frac{12-14}{2+2X_f} \] \[ -\frac23=\frac{-2}{2+2X_f} \] \(\textbf{Determine the required final resistance:}\) Cross-multiplying, \[ \frac23(2+2X_f)=2 \] \[ 2+2X_f=3.0 \] \[ X_f=0.50\,\Omega \] \(\textbf{Compare the power dissipated in X:}\) The current magnitude is identical before and after the change. Its resistance is different. Since \[ P_X=I_X^2X, \] the power ratio is \[ \frac{P_{X,f}}{P_{X,i}} =\frac{I_{X,f}^2X_f}{I_{X,i}^2X_i} \] The equal current magnitudes cancel: \[ \frac{P_{X,f}}{P_{X,i}}=\frac{X_f}{X_i} \] \[ \frac{P_{X,f}}{P_{X,i}}=\frac{0.50}{2.0} \] \[ \frac{P_{X,f}}{P_{X,i}}=\frac14 \] \(\textbf{Check the physical meaning:}\) Initially, the left source produces the larger clockwise driving EMF, giving downward shared current. After the right source is increased to \(14\,\text{V}\), its EMF exceeds the left-source EMF. The shared current therefore reverses when the resistance is suitably adjusted. Keeping the current magnitude unchanged does not keep the heating rate unchanged because the shared resistance is reduced. The two conditions—current reversal and equal current magnitude—must be imposed together to determine the new resistance. \(\textbf{Final answer:}\) \[ \boxed{X_f=0.50\,\Omega,\quad\frac{P_{X,f}}{P_{X,i}}=\frac14} \]

Explanation: \(\textbf{Apply the zero-current condition:}\) The actual current through the shared resistor is \[ I_s=i_L-i_R \] The ammeter shows \[ I_s=0 \] Therefore, \[ i_L=i_R \] The shared resistor has no potential drop because its current is zero. \(\textbf{Use the right-loop equation:}\) The right source supplies a clockwise potential rise of \(8.0\,\text{V}\). Since the shared branch has zero voltage drop, \[ 8.0=4.0i_R \] Thus, \[ i_R=2.0\,\text{A} \] \(\textbf{Determine the left mesh current:}\) The shared-current condition requires \[ i_L=i_R \] Hence, \[ i_L=2.0\,\text{A} \] \(\textbf{Recover the unknown resistance:}\) Kirchhoff's loop rule for the left mesh gives \[ 12=Xi_L+5.0(i_L-i_R) \] The shared-resistor term vanishes: \[ 12=X(2.0)+0 \] Therefore, \[ X=6.0\,\Omega \] \(\textbf{Calculate power supplied by the left source:}\) Its current passes through the source in the direction corresponding to electrical energy delivery. Thus, \[ P_L=\mathcal E_Li_L \] \[ P_L=(12)(2.0) \] \[ P_L=24\,\text{W} \] \(\textbf{Calculate power supplied by the right source:}\) \[ P_R=\mathcal E_Ri_R \] \[ P_R=(8.0)(2.0) \] \[ P_R=16\,\text{W} \] \(\textbf{Determine total source power:}\) \[ P_{\text{sources}}=P_L+P_R \] \[ P_{\text{sources}}=24+16 \] \[ P_{\text{sources}}=40\,\text{W} \] \(\textbf{Check resistive dissipation:}\) The left outer resistor dissipates \[ P_X=(2.0)^2(6.0)=24\,\text{W} \] The right outer resistor dissipates \[ P_4=(2.0)^2(4.0)=16\,\text{W} \] The shared resistor dissipates zero power. Total resistive dissipation is therefore \(40\,\text{W}\), consistent with the source powers. The zero shared current imposes a relation between the two mesh currents rather than requiring either mesh current to vanish. \(\textbf{Final answer:}\) \[ \boxed{X=6.0\,\Omega,\quad P_{\text{sources}}=40\,\text{W}} \]

Explanation: \(\textbf{Define the common-branch current:}\) Let the clockwise mesh currents be \(i_L\) and \(i_R\). The actual downward current through X is \[ I_X=i_L-i_R \] \(\textbf{Apply Kirchhoff's loop rule to the left loop:}\) The source provides a \(12\,\text{V}\) rise. The outer resistor and shared resistor produce the corresponding potential drops: \[ 12=2i_L+XI_X \] Therefore, \[ i_L=\frac{12-XI_X}{2} \] \(\textbf{Apply the right-loop equation:}\) The right mesh traverses X opposite to the downward reference direction. Thus, \[ 6.0=2i_R-XI_X \] Rearranging, \[ i_R=\frac{6.0+XI_X}{2} \] \(\textbf{Combine the two mesh relations:}\) \[ I_X=i_L-i_R \] Substituting, \[ I_X=\frac{12-XI_X}{2}-\frac{6.0+XI_X}{2} \] \[ 2I_X=6.0-2XI_X \] \[ (2+2X)I_X=6.0 \] Therefore, \[ I_X=\frac{3.0}{1+X} \] \(\textbf{Determine the initial current:}\) For \(X_i=2.0\,\Omega\), \[ I_{X,i}=\frac{3.0}{1+2.0} \] \[ I_{X,i}=1.0\,\text{A} \] The positive sign indicates downward current. \(\textbf{Initial internal heating in X:}\) \[ P_{X,i}=I_{X,i}^2X_i \] \[ P_{X,i}=(1.0)^2(2.0) \] \[ P_{X,i}=2.0\,\text{W} \] \(\textbf{Determine the changed current:}\) For \(X_f=5.0\,\Omega\), \[ I_{X,f}=\frac{3.0}{1+5.0} \] \[ I_{X,f}=0.50\,\text{A} \] The direction remains downwards. \(\textbf{Final heating rate:}\) \[ P_{X,f}=(0.50)^2(5.0) \] \[ P_{X,f}=1.25\,\text{W} \] \(\textbf{Compare the heating rates:}\) \[ \frac{P_{X,f}}{P_{X,i}}=\frac{1.25}{2.0} \] \[ \frac{P_{X,f}}{P_{X,i}}=\frac58 \] Increasing the shared resistance decreases its current. Although its resistance becomes greater, its heating rate decreases because the current reduction more than compensates. The current is determined by both coupled loops, not by treating X as though it were connected directly across either ideal source. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_{X,f}}{P_{X,i}}=\frac58,\quad I_{X,f}=0.50\,\text{A}\text{ downwards}} \]

Explanation: \(\textbf{Closed switch — Identify the effect of the ideal wire:}\) The common branch has zero resistance. The potential difference between its upper and lower ends is therefore zero. Each outer branch forms a closed path with this wire. \(\textbf{Current in the left outer branch:}\) The left source drives current through its \(2.0\,\Omega\) resistor: \[ i_L=\frac{12}{2.0} \] \[ i_L=6.0\,\text{A} \] \(\textbf{Current in the right outer branch:}\) \[ i_R=\frac{6.0}{2.0} \] \[ i_R=3.0\,\text{A} \] \(\textbf{Apply the current rule to the common wire:}\) Downward wire current is the algebraic difference of the clockwise mesh currents: \[ I_{\text{wire}}=i_L-i_R \] \[ I_{\text{wire}}=6.0-3.0 \] \[ I_{\text{wire}}=3.0\,\text{A} \] It flows downwards. \(\textbf{Determine source power with the switch closed:}\) Both sources supply electrical energy: \[ P_{\text{closed}}=12i_L+6.0i_R \] \[ P_{\text{closed}}=(12)(6.0)+(6.0)(3.0) \] \[ P_{\text{closed}}=72+18 \] \[ P_{\text{closed}}=90\,\text{W} \] \(\textbf{Open switch — Identify the remaining loop:}\) Opening the central branch leaves one closed conducting perimeter. The two ideal sources oppose one another around this outer loop. Their net driving EMF is \[ \mathcal E_{\text{net}}=12-6.0 \] \[ \mathcal E_{\text{net}}=6.0\,\text{V} \] The two outer resistors are now in series: \[ R_{\text{perimeter}}=2.0+2.0 \] \[ R_{\text{perimeter}}=4.0\,\Omega \] \(\textbf{Determine the perimeter current:}\) \[ I_{\text{open}}=\frac{6.0}{4.0} \] \[ I_{\text{open}}=1.50\,\text{A} \] \(\textbf{Calculate net source power with the switch open:}\) The \(12\,\text{V}\) source supplies energy, whereas current enters the positive terminal of the opposing \(6.0\,\text{V}\) source. Thus, \[ P_{\text{open}}=(12-6.0)(1.50) \] \[ P_{\text{open}}=9.0\,\text{W} \] \(\textbf{Compare net source powers:}\) \[ \frac{P_{\text{closed}}}{P_{\text{open}}} =\frac{90}{9.0} \] \[ \frac{P_{\text{closed}}}{P_{\text{open}}}=10 \] With the switch closed, each source has its own low-resistance loop, and both deliver electrical power. With the switch open, the sources oppose one another in a single perimeter loop. A switch in a shared branch can therefore change not only current magnitudes but also whether a source supplies or absorbs electrical energy. \(\textbf{Final answer:}\) \[ \boxed{I_{\text{wire}}=3.0\,\text{A}\text{ downwards},\quad\frac{P_{\text{closed}}}{P_{\text{open}}}=10} \]

Explanation: The Assertion is true because a mesh-current arrow is a chosen reference direction. It is not a physical constraint requiring current to flow along that arrow. Kirchhoff's equations describe the actual circuit independently of how the unknown current directions are assigned. A clockwise reference current of negative magnitude represents the same physical current as an anticlockwise reference current of equal positive magnitude. When the reference direction is changed, the signs of the corresponding potential-change terms must also be treated consistently. The current distribution obtained after solving the equations therefore remains physically unchanged. The Reason is true because electrical heating in an ohmic resistor follows the relation \(P=I^2R\). The sign of a current relative to a reference arrow disappears when its magnitude is squared. For a resistor shared by two meshes, the actual branch current must first be found from the algebraic contributions of both mesh currents. The same actual shared-branch current is obtained regardless of the chosen reference arrows. Consequently, the power in every resistor remains unchanged. The Reason correctly explains why changing a mathematical reference direction cannot change the physical energy dissipation of an unchanged circuit. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Initial circuit — Determine the series current:}\) Before the cell-containing branch is connected, the \(3.0\,\Omega\) and \(6.0\,\Omega\) resistors are in series. Their total resistance is \[ R_i=3.0+6.0 \] \[ R_i=9.0\,\Omega \] Therefore, \[ I_i=\frac{15}{9.0} \] \[ I_i=\frac53\,\text{A} \] \(\textbf{Initial junction potential:}\) The potential of B relative to D equals the drop across the \(6.0\,\Omega\) resistor: \[ V_{B,i}=I_i(6.0) \] \[ V_{B,i}=\frac53(6.0) \] \[ V_{B,i}=10\,\text{V} \] \(\textbf{Modified circuit — Assign the new junction potential:}\) Let the new potential of B be V. Current entering B through the \(3.0\,\Omega\) resistor is \[ I_{AB}=\frac{15-V}{3.0} \] Current leaving through the ordinary \(6.0\,\Omega\) resistor is \[ I_6=\frac V{6.0} \] \(\textbf{Express the current through the new cell branch:}\) Define positive current in this branch from B towards D. The cell's positive terminal is towards B. The branch voltage relation is \[ V=4.0+2.0I_{\text{cell}} \] Therefore, \[ I_{\text{cell}}=\frac{V-4.0}{2.0} \] \(\textbf{Apply the junction rule at B:}\) \[ I_{AB}=I_6+I_{\text{cell}} \] Substituting, \[ \frac{15-V}{3.0}=\frac V{6.0}+\frac{V-4.0}{2.0} \] Multiplying by six, \[ 2(15-V)=V+3(V-4.0) \] \[ 30-2V=4V-12 \] \[ 42=6V \] \[ V=7.0\,\text{V} \] \(\textbf{Determine the final source current:}\) \[ I_f=\frac{15-7.0}{3.0} \] \[ I_f=\frac83\,\text{A} \] Therefore, \[ \frac{I_f}{I_i}=\frac{8/3}{5/3} \] \[ \frac{I_f}{I_i}=\frac85 \] \(\textbf{Determine current through the rechargeable cell:}\) \[ I_{\text{cell}}=\frac{7.0-4.0}{2.0} \] \[ I_{\text{cell}}=1.50\,\text{A} \] This current enters the cell's positive terminal. The cell is therefore charging. \(\textbf{Calculate chemical-energy storage rate:}\) \[ P_{\text{chemical}}=\mathcal EI_{\text{ch}} \] \[ P_{\text{chemical}}=(4.0)(1.50) \] \[ P_{\text{chemical}}=6.0\,\text{W} \] \(\textbf{Check current conservation:}\) The ordinary \(6.0\,\Omega\) resistor carries \[ I_6=\frac76\,\text{A} \] The outgoing currents add to \[ \frac76+\frac32=\frac83\,\text{A} \] This agrees with the source current. Connecting the additional branch lowers the potential at B and increases source current. The new cell nevertheless receives charging current because its terminal voltage exceeds its EMF. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_f}{I_i}=\frac85,\quad P_{\text{chemical}}=6.0\,\text{W}} \]

Explanation: \(\textbf{Determine the initial shared-branch current:}\) Taking downward current as positive, \[ I_{X,i}=i_L-i_R \] \[ I_{X,i}=4.0-3.0 \] \[ I_{X,i}=1.0\,\text{A} \] \(\textbf{Recover X from the left loop:}\) Kirchhoff's loop rule gives \[ 14=3.0i_L+XI_{X,i} \] Substituting, \[ 14=3.0(4.0)+X(1.0) \] \[ 14=12+X \] \[ X=2.0\,\Omega \] \(\textbf{Determine the original right-cell EMF:}\) The right clockwise loop equation is \[ E=2.0i_R+X(i_R-i_L) \] Substituting, \[ E=2.0(3.0)+2.0(3.0-4.0) \] \[ E=6.0-2.0 \] \[ E=4.0\,\text{V} \] \(\textbf{Construct the loop equations after reversal:}\) The right cell now gives a clockwise potential drop of \(4.0\,\text{V}\). Let the new clockwise mesh currents be \(j_L\) and \(j_R\). The left loop gives \[ 14=3j_L+2(j_L-j_R) \] \[ 5j_L-2j_R=14 \] The right loop gives \[ -4.0=2j_R+2(j_R-j_L) \] \[ -2j_L+4j_R=-4.0 \] \(\textbf{Solve the changed mesh currents:}\) Divide the second equation by two: \[ -j_L+2j_R=-2.0 \] Thus, \[ 2j_R=j_L-2.0 \] Substituting into the left equation, \[ 5j_L-(j_L-2.0)=14 \] \[ 4j_L=12 \] \[ j_L=3.0\,\text{A} \] Therefore, \[ 2j_R=3.0-2.0 \] \[ j_R=0.50\,\text{A} \] \(\textbf{Determine the new shared current:}\) \[ I_{X,f}=j_L-j_R \] \[ I_{X,f}=3.0-0.50 \] \[ I_{X,f}=2.50\,\text{A} \] \(\textbf{Compare shared-resistor heating:}\) The resistance X is unchanged. Therefore, \[ \frac{P_{X,f}}{P_{X,i}} =\left(\frac{I_{X,f}}{I_{X,i}}\right)^2 \] \[ \frac{P_{X,f}}{P_{X,i}} =\left(\frac{2.50}{1.0}\right)^2 \] \[ \frac{P_{X,f}}{P_{X,i}}=\frac{25}{4} \] The initial mesh-current measurements determine both the shared resistance and the unknown right-cell EMF. Reversing that source then changes the current distribution between the coupled loops. The shared current increases substantially, even though the resistance of its branch is unchanged. \(\textbf{Final answer:}\) \[ \boxed{X=2.0\,\Omega,\quad\frac{P_{X,f}}{P_{X,i}}=\frac{25}{4}} \]

Explanation: \(\textbf{Identify Q's terminal-voltage relation:}\) The current enters Q's positive terminal. Therefore, Q is charging. For a charging real cell, \[ V_Q=\mathcal E_Q+Ir_Q \] The voltmeter reads \(9.0\,\text{V}\), and \(r_Q=1.0\,\Omega\). Thus, \[ 9.0=\mathcal E_Q+I(1.0) \] Rearranging, \[ \mathcal E_Q=9.0-I \] \(\textbf{Determine the complete series resistance:}\) Both cell internal resistances and the external resistor carry the same current: \[ R_{\text{total}}=1.0+1.0+3.0 \] \[ R_{\text{total}}=5.0\,\Omega \] \(\textbf{Apply Kirchhoff's loop rule:}\) The resultant driving EMF is the difference between the opposing EMFs. Therefore, \[ 17-\mathcal E_Q=5.0I \] \(\textbf{Combine the cell-voltage and loop equations:}\) Substitute \(\mathcal E_Q=9.0-I\): \[ 17-(9.0-I)=5.0I \] \[ 8.0+I=5.0I \] \[ 4.0I=8.0 \] \[ I=2.0\,\text{A} \] \(\textbf{Recover Q's EMF:}\) \[ \mathcal E_Q=9.0-2.0 \] \[ \mathcal E_Q=7.0\,\text{V} \] \(\textbf{Determine external resistor power:}\) \[ P_R=I^2R \] \[ P_R=(2.0)^2(3.0) \] \[ P_R=12\,\text{W} \] \(\textbf{Check the loop potential balance:}\) The resultant EMF is \[ 17-7.0=10\,\text{V} \] The total resistive drop is \[ I R_{\text{total}}=(2.0)(5.0)=10\,\text{V} \] The loop rule is satisfied. \(\textbf{Energy-conservation check:}\) P generates \[ P_P=(17)(2.0)=34\,\text{W} \] Q stores chemical energy at the rate \[ P_{\text{chemical},Q}=(7.0)(2.0)=14\,\text{W} \] The two internal resistances dissipate \(8.0\,\text{W}\) altogether, while the external resistor dissipates \(12\,\text{W}\). Thus, \[ 14+8.0+12=34\,\text{W} \] The measured \(9.0\,\text{V}\) across Q exceeds its EMF because charging current produces an additional internal resistive voltage drop. The voltmeter reading cannot be substituted directly for Q's EMF in the complete loop equation. \(\textbf{Final answer:}\) \[ \boxed{\mathcal E_Q=7.0\,\text{V},\quad P_R=12\,\text{W}} \]

Explanation: For observation P, current leaves the positive terminal of the ideal cell. The cell is therefore delivering electrical energy to the external circuit. Its delivered power is the product of its \(8.0\,\text{V}\) EMF and \(2.0\,\text{A}\) current. \[ P_P=\mathcal EI=(8.0)(2.0)=16\,\text{W} \] Thus, P matches 2. \(\textbf{Observation Q — an ideal cell receiving current:}\) For observation Q, current enters the positive terminal of the ideal rechargeable cell. The cell consequently absorbs electrical energy rather than supplying it. \[ P_Q=(8.0)(1.0)=8.0\,\text{W} \] Thus, Q matches 1. \(\textbf{Observation R — resistive heating:}\) For observation R, the potential difference and resistance determine the current: \[ I_R=\frac{8.0}{4.0}=2.0\,\text{A} \] The resistor dissipates \[ P_R=VI=(8.0)(2.0)=16\,\text{W} \] It does not store that energy chemically. Thus, R matches 3. \(\textbf{Observation S — charging a real cell:}\) For observation S, the measured terminal potential difference exceeds the cell EMF. The real cell must therefore be in the charging condition under the stated model. Its charging-current magnitude is \[ I_S=\frac{8.0-6.0}{2.0}=1.0\,\text{A} \] The chemical-energy storage rate is \[ P_{\text{chemical}}=\mathcal EI=(6.0)(1.0)=6.0\,\text{W} \] The remaining \(2.0\,\text{W}\) of the terminal electrical input is dissipated in the internal resistance. Thus, S matches 4. The distinction is between electrical energy generated by a source, electrical energy absorbed through terminals, resistive heat and chemical-energy storage. The complete matching is P–2, Q–1, R–3 and S–4.

Explanation: \(\textbf{Define the common junction voltage:}\) Let \[ V=V_U-V_L \] The algebraic currents supplied upwards by the cells are \[ I_P=\frac{12-V}{2.0} \] \[ I_Q=\frac{6.0-V}{2.0} \] The external resistor carries downward current \[ I_R=\frac V{6.0} \] \(\textbf{Include the independently supplied current:}\) The ideal current source injects J into U. Kirchhoff's junction rule gives \[ I_P+I_Q+J=I_R \] Therefore, \[ \frac{12-V}{2.0}+\frac{6.0-V}{2.0}+J=\frac V{6.0} \] Combining, \[ 9.0-V+J=\frac V6 \] \(\textbf{Obtain the terminal-voltage relation:}\) Multiplying by six, \[ 54-6V+6J=V \] \[ 7V=54+6J \] Hence, \[ V=\frac{54+6J}{7} \] \(\textbf{Determine the zero-current condition for P:}\) Cell P carries zero current when its terminal voltage equals its EMF: \[ I_P=0 \] \[ V=12\,\text{V} \] Substitute this voltage into the junction relation: \[ 12=\frac{54+6J}{7} \] \[ 84=54+6J \] \[ 6J=30 \] \[ J=5.0\,\text{A} \] This is the threshold at which P changes from supplying current to receiving charging current. \(\textbf{Determine the terminal voltage when }J=6.0\,\textbf{A:}\) \[ V=\frac{54+6(6.0)}7 \] \[ V=\frac{90}{7}\,\text{V} \] This exceeds P's EMF of \(12\,\text{V}\). \(\textbf{Calculate P's algebraic supplying current:}\) \[ I_P=\frac{12-90/7}{2.0} \] \[ I_P=\frac{84/7-90/7}{2.0} \] \[ I_P=-\frac37\,\text{A} \] The negative sign indicates current entering P's positive terminal. Thus, its charging-current magnitude is \[ I_{\text{ch},P}=\frac37\,\text{A} \] \(\textbf{Check current conservation:}\) At this setting, Q also receives charging current: \[ I_Q=\frac{6.0-90/7}{2.0} \] \[ I_Q=-\frac{24}{7}\,\text{A} \] The external resistor carries \[ I_R=\frac{90/7}{6.0} \] \[ I_R=\frac{15}{7}\,\text{A} \] The injected current is \[ J=\frac{42}{7}\,\text{A} \] The currents leaving U total \[ \frac37+\frac{24}{7}+\frac{15}{7} \] \[ \frac{42}{7}=6.0\,\text{A} \] The junction rule is satisfied. An independent current source can raise the common terminal voltage enough to reverse the operating direction of a cell without changing its polarity or EMF. The transition occurs when its branch current becomes zero. \(\textbf{Final answer:}\) \[ \boxed{J_{\text{threshold}}=5.0\,\text{A},\quad I_{\text{ch},P}=\frac37\,\text{A}} \]

Explanation: \(\textbf{Reduce the original parallel section:}\) The \(3.0\,\Omega\) and \(6.0\,\Omega\) resistors are initially parallel. Their equivalent resistance is \[ R_{p,i}=\frac{(3.0)(6.0)}{3.0+6.0} \] \[ R_{p,i}=2.0\,\Omega \] \(\textbf{Determine the original source current:}\) The parallel section is in series with P: \[ R_i=2.0+2.0 \] \[ R_i=4.0\,\Omega \] Therefore, \[ I_i=\frac{30}{4.0} \] \[ I_i=7.50\,\text{A} \] \(\textbf{Determine the initial parallel-section voltage:}\) \[ V_{p,i}=I_iR_{p,i} \] \[ V_{p,i}=(7.50)(2.0) \] \[ V_{p,i}=15\,\text{V} \] \(\textbf{Infer equivalent resistance from the faulty-circuit current:}\) The source still maintains \(30\,\text{V}\). Its final current is \(6.0\,\text{A}\). Therefore, \[ R_f=\frac{30}{6.0} \] \[ R_f=5.0\,\Omega \] \(\textbf{Identify the surviving branch resistance:}\) P remains connected in series and has resistance \(2.0\,\Omega\). Therefore, the resistance of the surviving parallel-section path is \[ R_{\text{surviving}}=R_f-R_P \] \[ R_{\text{surviving}}=5.0-2.0 \] \[ R_{\text{surviving}}=3.0\,\Omega \] Thus, the \(3.0\,\Omega\) branch survives. The \(6.0\,\Omega\) branch has failed open. \(\textbf{Determine the new voltage across the surviving resistor:}\) \[ V_{p,f}=I_fR_{\text{surviving}} \] \[ V_{p,f}=(6.0)(3.0) \] \[ V_{p,f}=18\,\text{V} \] \(\textbf{Compare its electrical powers:}\) The surviving resistor is unchanged. Its power follows \[ P=\frac{V^2}{R} \] Therefore, \[ \frac{P_f}{P_i}=\left(\frac{V_{p,f}}{V_{p,i}}\right)^2 \] \[ \frac{P_f}{P_i}=\left(\frac{18}{15}\right)^2 \] \[ \frac{P_f}{P_i}=\frac{36}{25} \] \(\textbf{Interpret the apparent power increase:}\) The source current decreases because the failed branch raises total equivalent resistance. However, the voltage drop across P decreases. A greater fraction of source voltage therefore appears across the surviving parallel resistor. Its individual power increases despite the reduction in total source current. The source-current measurement identifies which branch failed only after the changed series-parallel topology is considered. \(\textbf{Final answer:}\) The failed branch is the \(6.0\,\Omega\) branch. The ratio of final to initial power dissipated in the surviving \(3.0\,\Omega\) resistor is \[ \frac{P_f}{P_i}=\frac{36}{25} \]

Explanation: \(\textbf{Define the unknown junction potential:}\) Let \[ V_B=V \] The source fixes the potential of A at \(V_s\), while D remains at zero. \(\textbf{Write the current entering B from A:}\) \[ I_{AB}=\frac{V_s-V}{2.0} \] \(\textbf{Write the current leaving through the ordinary resistor:}\) \[ I_4=\frac V{4.0} \] \(\textbf{Write the current through the cell branch:}\) Take current from B towards D as positive. The cell's positive terminal is towards B. Thus, \[ V=6.0+2.0I_{\text{cell}} \] Therefore, \[ I_{\text{cell}}=\frac{V-6.0}{2.0} \] \(\textbf{Apply Kirchhoff's junction rule:}\) \[ I_{AB}=I_4+I_{\text{cell}} \] Substituting, \[ \frac{V_s-V}{2.0}=\frac V{4.0}+\frac{V-6.0}{2.0} \] Multiplying by four, \[ 2(V_s-V)=V+2(V-6.0) \] \[ 2V_s-2V=3V-12 \] \[ 5V=2V_s+12 \] Therefore, \[ V=\frac{2V_s+12}{5} \] \(\textbf{Determine the zero-current threshold:}\) The cell carries zero current when the potential across its external terminals equals its EMF: \[ I_{\text{cell}}=0 \] \[ V=6.0\,\text{V} \] Substituting into the junction-potential relation, \[ 6.0=\frac{2V_s+12}{5} \] \[ 30=2V_s+12 \] \[ V_s=9.0\,\text{V} \] \(\textbf{Determine junction voltage at the final source setting:}\) For \(V_s=14\,\text{V}\), \[ V=\frac{2(14)+12}{5} \] \[ V=8.0\,\text{V} \] \(\textbf{Determine the cell-branch current:}\) \[ I_{\text{cell}}=\frac{8.0-6.0}{2.0} \] \[ I_{\text{cell}}=1.0\,\text{A} \] Current enters the cell's positive terminal, so it is charging. \(\textbf{Calculate chemical-energy storage rate:}\) \[ P_{\text{chemical}}=\mathcal EI_{\text{ch}} \] \[ P_{\text{chemical}}=(6.0)(1.0) \] \[ P_{\text{chemical}}=6.0\,\text{W} \] \(\textbf{Verify current conservation:}\) At the final setting, \[ I_{AB}=\frac{14-8.0}{2.0}=3.0\,\text{A} \] The ordinary resistor carries \[ I_4=\frac{8.0}{4.0}=2.0\,\text{A} \] The remaining \(1.0\,\text{A}\) charges the cell. Increasing the control voltage changes the potential of B continuously. The cell changes from discharging to charging when that junction potential passes through its EMF. No reversal of the cell itself is required. \(\textbf{Final answer:}\) \[ \boxed{V_{s,\text{threshold}}=9.0\,\text{V},\quad P_{\text{chemical}}=6.0\,\text{W}} \]

Explanation: Statement I is correct because the potential change across each circuit element depends on the direction in which that element is traversed. Reversing the traversal direction changes the sign of every corresponding potential change. The resulting loop equation is algebraically equivalent to the original equation because both express a zero net potential change around the same closed loop. The actual branch currents and voltages remain unchanged. Statement II is incorrect because a negative mesh-current value only indicates that the actual current is opposite to the chosen mesh-current arrow. Whether a source supplies or absorbs power depends on the actual current direction relative to its positive and negative terminals. A source can deliver power even when a mesh current defined clockwise is negative. Statement III is correct because mesh currents are mathematical current contributions assigned to closed paths. A resistor common to two meshes carries the algebraic sum of the contributions through that shared branch. When the two mesh directions traverse it oppositely, the actual branch current is their algebraic difference. Using either mesh-current magnitude alone would generally give the wrong resistor power. Statement IV is correct because Kirchhoff's junction rule includes every branch meeting at the junction. If a set of measured outgoing currents exceeds a set of measured incoming currents, a remaining branch can supply the difference. The complete algebraic current balance must be zero. There is no requirement for each partial group of branch currents to balance separately. Statements I, III and IV are correct.