Current Electricity MCQs with Answers – Part 3 (NEET Physics)

Explanation: \(\textbf{Reduce the upper branch:}\) The two \(6.0\,\Omega\) resistors in the upper path are in series. Therefore, \[ R_{\text{upper}}=6.0+6.0 \] \[ R_{\text{upper}}=12\,\Omega \] \(\textbf{Combine the parallel paths:}\) The lower branch also has resistance \(12\,\Omega\). Thus, the two B–C paths have equal resistance. Their equivalent is \[ R_{BC}=\frac{(12)(12)}{12+12} \] \[ R_{BC}=6.0\,\Omega \] \(\textbf{Determine total circuit resistance:}\) The two outer \(3.0\,\Omega\) resistors are in series with the parallel section. \[ R_{\text{eq}}=3.0+6.0+3.0 \] \[ R_{\text{eq}}=12\,\Omega \] \(\textbf{Source current:}\) \[ I_{\text{source}}=\frac{24}{12} \] \[ I_{\text{source}}=2.0\,\text{A} \] \(\textbf{Determine branch currents:}\) The two parallel B–C paths have equal resistance. Therefore, the source current divides equally. \[ I_{\text{upper}}=I_{\text{lower}}=\frac{2.0}{2} \] \[ I_{\text{upper}}=I_{\text{lower}}=1.0\,\text{A} \] The same \(1.0\,\text{A}\) passes through both series resistors in the upper branch. \(\textbf{Power in the lower branch:}\) \[ P_{\text{lower}}=I_{\text{lower}}^2R_{\text{lower}} \] \[ P_{\text{lower}}=(1.0)^2(12) \] \[ P_{\text{lower}}=12\,\text{W} \] \(\textbf{Power in one upper resistor:}\) \[ P_{6}=(1.0)^2(6.0) \] \[ P_6=6.0\,\text{W} \] Therefore, \[ \frac{P_{\text{lower}}}{P_6}=\frac{12}{6.0} \] \[ \frac{P_{\text{lower}}}{P_6}=2 \] The complete upper and lower parallel paths dissipate equal total powers because their resistances and applied voltages are equal. However, the upper path's power is shared between two series resistors. The single lower resistor therefore dissipates twice as much power as either individual upper resistor. The distinction between an entire branch and a component within that branch is essential. \(\textbf{Final answer:}\) \[ \boxed{I_{\text{source}}=2.0\,\text{A},\quad\frac{P_{\text{lower}}}{P_6}=2} \]

Explanation: \(\textbf{Identify the common series current:}\) The resistors are connected in series. Therefore, the same current passes through each. Their individual power dissipation is \[ P=I^2R \] Each resistor places an independent upper limit on the permitted current. \(\textbf{Current limit imposed by the }3.0\,\Omega\textbf{ resistor:}\) Its maximum safe power is \[ P_{3,\max}=12\,\text{W} \] Therefore, \[ I_{3,\max}=\sqrt{\frac{P_{3,\max}}{R_3}} \] \[ I_{3,\max}=\sqrt{\frac{12}{3.0}} \] \[ I_{3,\max}=2.0\,\text{A} \] \(\textbf{Current limit imposed by the }6.0\,\Omega\textbf{ resistor:}\) Its maximum safe power is \[ P_{6,\max}=6.0\,\text{W} \] Hence, \[ I_{6,\max}=\sqrt{\frac{6.0}{6.0}} \] \[ I_{6,\max}=1.0\,\text{A} \] \(\textbf{Select the actual circuit-current limit:}\) The same current flows through both resistors. Therefore, the circuit must satisfy the more restrictive limit: \[ I_{\max}=1.0\,\text{A} \] The \(6.0\,\Omega\) resistor reaches its safe power limit first. \(\textbf{Determine the total series resistance:}\) \[ R_{\text{eq}}=3.0+6.0 \] \[ R_{\text{eq}}=9.0\,\Omega \] \(\textbf{Maximum safe source voltage:}\) \[ V_{\max}=I_{\max}R_{\text{eq}} \] \[ V_{\max}=(1.0)(9.0) \] \[ V_{\max}=9.0\,\text{V} \] \(\textbf{Power in the other resistor at this voltage:}\) \[ P_3=I_{\max}^2R_3 \] \[ P_3=(1.0)^2(3.0) \] \[ P_3=3.0\,\text{W} \] Although the \(3.0\,\Omega\) resistor can tolerate \(12\,\text{W}\), the complete circuit cannot be operated at the corresponding \(2.0\,\text{A}\). At that current, the \(6.0\,\Omega\) resistor would dissipate \(24\,\text{W}\), exceeding its \(6.0\,\text{W}\) limit. The greatest safe source voltage is determined by the first component to reach its rated dissipation. \(\textbf{Final answer:}\) \[ \boxed{V_{\max}=9.0\,\text{V},\quad P_3=3.0\,\text{W}} \]

Explanation: \(\textbf{Initial resistance and current in branch P:}\) The two resistors are in series. \[ R_{P,i}=4.0+6.0 \] \[ R_{P,i}=10\,\Omega \] Since P is connected directly across the ideal source, \[ I_{P,i}=\frac{20}{10} \] \[ I_{P,i}=2.0\,\text{A} \] \(\textbf{Initial power in the unchanged }6.0\,\Omega\textbf{ resistor:}\) \[ P_{6,i}=I_{P,i}^2R \] \[ P_{6,i}=(2.0)^2(6.0) \] \[ P_{6,i}=24\,\text{W} \] \(\textbf{Analyse branch Q:}\) Branch Q contains two \(5.0\,\Omega\) resistors in series. \[ R_Q=5.0+5.0 \] \[ R_Q=10\,\Omega \] Its current is \[ I_Q=\frac{20}{10} \] \[ I_Q=2.0\,\text{A} \] This current remains unchanged because the ideal source voltage and branch Q's resistance do not change. \(\textbf{Changed resistance in branch P:}\) The original \(4.0\,\Omega\) resistor becomes \(6.0\,\Omega\). Therefore, \[ R_{P,f}=6.0+6.0 \] \[ R_{P,f}=12\,\Omega \] \(\textbf{New current in branch P:}\) \[ I_{P,f}=\frac{20}{12} \] \[ I_{P,f}=\frac53\,\text{A} \] \(\textbf{Final total source current:}\) The two branches remain in parallel. Therefore, \[ I_{\text{source},f}=I_{P,f}+I_Q \] \[ I_{\text{source},f}=\frac53+2 \] \[ I_{\text{source},f}=\frac{11}{3}\,\text{A} \] \(\textbf{New power in the original }6.0\,\Omega\textbf{ resistor:}\) The resistor's resistance remains \(6.0\,\Omega\), but its current changes. Therefore, \[ P_{6,f}=\left(\frac53\right)^2(6.0) \] \[ P_{6,f}=\frac{25}{9}(6.0) \] \[ P_{6,f}=\frac{50}{3}\,\text{W} \] \(\textbf{Compare the powers:}\) \[ \frac{P_{6,f}}{P_{6,i}} =\frac{50/3}{24} \] \[ \frac{P_{6,f}}{P_{6,i}}=\frac{25}{36} \] The increase in resistance affects the current throughout branch P because its two resistors are in series. However, branch Q remains directly connected across the same ideal voltage and therefore carries its original current. The power of the unchanged resistor in P decreases because its branch current decreases. An individual resistor's power can change even when its own resistance does not. \(\textbf{Final answer:}\) \[ \boxed{I_{\text{source},f}=\frac{11}{3}\,\text{A},\quad\frac{P_{6,f}}{P_{6,i}}=\frac{25}{36}} \]

Explanation: \(\textbf{Resistance of each cut piece:}\) The original wire has resistance \[ R_0=18\,\Omega \] For a uniform wire, \[ R=\frac{\rho L}{A} \] Cutting the wire into three equal lengths leaves the resistivity and cross-sectional area unchanged. Therefore, each piece has one-third of the original resistance: \[ R_P=R_Q=R_R=\frac{18}{3} \] \[ R_P=R_Q=R_R=6.0\,\Omega \] \(\textbf{Reduce the parallel section:}\) Pieces Q and R are connected in parallel. \[ \frac1{R_{QR}}=\frac16+\frac16 \] \[ \frac1{R_{QR}}=\frac13 \] \[ R_{QR}=3.0\,\Omega \] \(\textbf{Determine the complete resistance:}\) Piece P is in series with this parallel combination. \[ R_{\text{eq}}=R_P+R_{QR} \] \[ R_{\text{eq}}=6.0+3.0 \] \[ R_{\text{eq}}=9.0\,\Omega \] \(\textbf{Determine the voltage across Q:}\) The source voltage is \[ V=18\,\text{V} \] The total current is \[ I=\frac{18}{9.0} \] \[ I=2.0\,\text{A} \] The potential difference across the parallel section is \[ V_{QR}=IR_{QR} \] \[ V_{QR}=(2.0)(3.0) \] \[ V_{QR}=6.0\,\text{V} \] Piece Q experiences this same voltage because it is one of the parallel branches. \(\textbf{Power dissipated in piece Q:}\) \[ P_Q=\frac{V_{QR}^2}{R_Q} \] \[ P_Q=\frac{6.0^2}{6.0} \] \[ P_Q=6.0\,\text{W} \] \(\textbf{Compare with the original uncut wire:}\) At \(18\,\text{V}\), the original \(18\,\Omega\) wire would dissipate \[ P_0=\frac{18^2}{18} \] \[ P_0=18\,\text{W} \] Therefore, \[ \frac{P_Q}{P_0}=\frac{6.0}{18} \] \[ \frac{P_Q}{P_0}=\frac13 \] The total electrical power drawn by the rearranged pieces is greater than that drawn by the uncut wire because the equivalent resistance is smaller. However, piece Q alone dissipates only one-third of the original wire's power. Cutting a wire does not change its material resistivity, but the electrical arrangement of the pieces determines the resulting resistance and distribution of power. \(\textbf{Final answer:}\) \[ \boxed{R_{\text{eq}}=9.0\,\Omega,\quad\frac{P_Q}{P_0}=\frac13} \]

Explanation: \(\textbf{Equivalent resistance of arrangement P:}\) The \(3.0\,\Omega\) and \(6.0\,\Omega\) resistors are in parallel. \[ R_{p,P}=\frac{(3.0)(6.0)}{3.0+6.0} \] \[ R_{p,P}=2.0\,\Omega \] The \(2.0\,\Omega\) series resistor adds to this combination. \[ R_P=2.0+2.0 \] \[ R_P=4.0\,\Omega \] \(\textbf{Source current in P:}\) \[ I_P=\frac{12}{4.0} \] \[ I_P=3.0\,\text{A} \] The potential difference across its parallel section is \[ V_{p,P}=(3.0)(2.0) \] \[ V_{p,P}=6.0\,\text{V} \] Therefore, the \(6.0\,\Omega\) resistor in arrangement P dissipates \[ P_{6,P}=\frac{6.0^2}{6.0} \] \[ P_{6,P}=6.0\,\text{W} \] \(\textbf{Equivalent resistance of arrangement Q:}\) The \(2.0\,\Omega\) and \(3.0\,\Omega\) resistors are in parallel. \[ R_{p,Q}=\frac{(2.0)(3.0)}{2.0+3.0} \] \[ R_{p,Q}=\frac65\,\Omega \] The \(6.0\,\Omega\) resistor is in series with this combination. \[ R_Q=6.0+\frac65 \] \[ R_Q=\frac{36}{5}\,\Omega \] \(\textbf{Source current in Q:}\) \[ I_Q=\frac{12}{36/5} \] \[ I_Q=\frac53\,\text{A} \] Hence, \[ \frac{I_P}{I_Q}=\frac{3.0}{5/3} \] \[ \frac{I_P}{I_Q}=\frac95 \] \(\textbf{Power in the }6.0\,\Omega\textbf{ resistor in Q:}\) The \(6.0\,\Omega\) resistor now carries the complete source current. \[ P_{6,Q}=I_Q^2(6.0) \] \[ P_{6,Q}=\left(\frac53\right)^2(6.0) \] \[ P_{6,Q}=\frac{50}{3}\,\text{W} \] Therefore, \[ \frac{P_{6,Q}}{P_{6,P}} =\frac{50/3}{6.0} \] \[ \frac{P_{6,Q}}{P_{6,P}}=\frac{25}{9} \] The total source current is greater in P because its equivalent resistance is smaller. Nevertheless, the \(6.0\,\Omega\) resistor dissipates more power in Q because it carries the entire series current instead of only a fraction of the current entering a parallel section. The current through an individual component cannot be inferred from total source current without considering its position in the network. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_P}{I_Q}=\frac95,\quad\frac{P_{6,Q}}{P_{6,P}}=\frac{25}{9}} \]

Explanation: \(\textbf{Relate source power to equivalent resistance:}\) For an ideal constant-voltage source supplying ohmic resistors, \[ P=\frac{V^2}{R_{\text{eq}}} \] Both experiments use the same applied voltage. Therefore, the ratio of source powers is the inverse ratio of their equivalent resistances. \(\textbf{Series-equivalent resistance:}\) The given sum is \[ R_1+R_2=18\,\Omega \] Hence, \[ R_s=18\,\Omega \] \(\textbf{Parallel-equivalent resistance:}\) For two resistors, \[ R_p=\frac{R_1R_2}{R_1+R_2} \] The power ratio is \[ \frac{P_Q}{P_P}=\frac{R_s}{R_p} \] Therefore, \[ \frac92=\frac{18}{R_p} \] Rearranging, \[ R_p=18\left(\frac29\right) \] \[ R_p=4.0\,\Omega \] \(\textbf{Infer the product of the unknown resistances:}\) Using the parallel relation, \[ 4.0=\frac{R_1R_2}{18} \] Therefore, \[ R_1R_2=72\,\Omega^2 \] \(\textbf{Determine the individual resistances:}\) The resistances have sum \(18\) and product \(72\). Thus, their values are the roots of \[ x^2-18x+72=0 \] Factorising, \[ (x-6)(x-12)=0 \] Therefore, \[ R_{\text{small}}=6.0\,\Omega \] \[ R_{\text{large}}=12\,\Omega \] \(\textbf{Power in the larger resistor during parallel operation:}\) In experiment Q, both resistors experience the full \(18\,\text{V}\). For the larger resistor, \[ P_{\text{large},Q}=\frac{18^2}{12} \] \[ P_{\text{large},Q}=27\,\text{W} \] \(\textbf{Power-balance check:}\) The \(6.0\,\Omega\) resistor dissipates \[ P_{\text{small},Q}=\frac{18^2}{6.0} \] \[ P_{\text{small},Q}=54\,\text{W} \] The total parallel power is \(81\,\text{W}\). The series power is \[ P_P=\frac{18^2}{18}=18\,\text{W} \] Their ratio is \[ \frac{81}{18}=\frac92 \] This verifies the supplied observation. A comparison of total powers in two arrangements reveals the parallel-equivalent resistance, which can be combined with the known resistance sum to infer both individual values. \(\textbf{Final answer:}\) \[ \boxed{R_{\text{large}}=12\,\Omega,\quad P_{\text{large},Q}=27\,\text{W}} \]

Explanation: \(\textbf{Recognise the changed branch:}\) Initially, the three resistors form separate parallel paths. The \(4.0\,\Omega\) resistor is inserted only into the branch that originally contained \(2.0\,\Omega\). The other branch resistances remain \(3.0\,\Omega\) and \(6.0\,\Omega\). \(\textbf{Initial equivalent resistance:}\) \[ \frac1{R_i}=\frac13+\frac16+\frac12 \] \[ \frac1{R_i}=1 \] \[ R_i=1.0\,\Omega \] The original source voltage is consequently \[ V_i=IR_i \] \[ V_i=(6.0)(1.0) \] \[ V_i=6.0\,\text{V} \] \(\textbf{Resistance of the modified branch:}\) The original \(2.0\,\Omega\) resistor and the added \(4.0\,\Omega\) resistor are in series. \[ R_{\text{modified}}=2.0+4.0 \] \[ R_{\text{modified}}=6.0\,\Omega \] The three branch resistances are now \(3.0\,\Omega\), \(6.0\,\Omega\) and \(6.0\,\Omega\). \(\textbf{Final equivalent resistance:}\) \[ \frac1{R_f}=\frac13+\frac16+\frac16 \] \[ \frac1{R_f}=\frac23 \] \[ R_f=\frac32\,\Omega \] \(\textbf{Determine the new network voltage:}\) The ideal current source maintains its original total current. \[ V_f=IR_f \] \[ V_f=(6.0)\left(\frac32\right) \] \[ V_f=9.0\,\text{V} \] \(\textbf{Current through the modified branch:}\) The modified branch experiences the complete parallel-network voltage. \[ I_{\text{modified}}=\frac{V_f}{R_{\text{modified}}} \] \[ I_{\text{modified}}=\frac{9.0}{6.0} \] \[ I_{\text{modified}}=1.50\,\text{A} \] Both resistors in that branch carry this current. \(\textbf{Power in the added resistor:}\) \[ P_4=I_{\text{modified}}^2(4.0) \] \[ P_4=(1.50)^2(4.0) \] \[ P_4=9.0\,\text{W} \] \(\textbf{Current-conservation check:}\) The final currents through the other branches are \[ I_3=\frac{9.0}{3.0}=3.0\,\text{A} \] \[ I_6=\frac{9.0}{6.0}=1.50\,\text{A} \] Adding all three branch currents, \[ 3.0+1.50+1.50=6.0\,\text{A} \] This agrees with the maintained source current. Inserting a series resistor into one parallel branch raises the network's equivalent resistance. For an ideal constant-current source, the terminal voltage must rise to maintain the total current. As a result, currents in the unchanged branches can increase even though another branch has acquired additional resistance. \(\textbf{Final answer:}\) \[ \boxed{V_f=9.0\,\text{V},\quad P_4=9.0\,\text{W}} \]

Explanation: Statement I is correct because adding a positive series resistance increases the equivalent resistance of the complete conducting path. When an ideal source maintains a fixed voltage, the circuit current follows \(I=V/R_{\text{eq}}\). An increase in equivalent resistance therefore reduces the source current. Statement II is incorrect because the existing branches remain connected directly across the same ideal source. Their potential differences consequently remain unchanged. If their resistance values also remain unchanged, their individual currents remain unchanged. The source supplies additional current to the newly connected branch, but this does not require the original branch currents to change. Statement III is correct because an ideal constant-current source maintains total current rather than a fixed terminal voltage. Adding a parallel conducting branch reduces the equivalent resistance of a positive-resistance parallel network. The terminal voltage follows \(V=I_{\text{source}}R_{\text{eq}}\), so it decreases when the equivalent resistance decreases under fixed source current. Statement IV is correct because resistors in series carry the same current. Their powers are given by \(P_i=I^2R_i\). Thus, for the same series current, the larger resistance dissipates more electrical power. The conclusions for the two source types must not be interchanged. A fixed-voltage source responds to a resistance change by changing total current, whereas a fixed-current source responds by changing terminal voltage. Likewise, equal voltage in parallel and equal current in series lead to different power-sharing rules. Therefore, Statements I, III and IV are correct.

Explanation: \(\textbf{Determine the paths in experiment P:}\) The source is connected between A and B. There are two conducting paths between these terminals. The direct A–B path contains the \(3.0\,\Omega\) resistor. The indirect path A–C–B contains the \(9.0\,\Omega\) and \(6.0\,\Omega\) resistors in series. Therefore, \[ R_{\text{indirect},P}=9.0+6.0 \] \[ R_{\text{indirect},P}=15\,\Omega \] The two paths are parallel. \(\textbf{Source current in experiment P:}\) The direct-branch current is \[ I_{AB,P}=\frac{18}{3.0} \] \[ I_{AB,P}=6.0\,\text{A} \] The indirect-branch current is \[ I_{\text{indirect},P}=\frac{18}{15} \] \[ I_{\text{indirect},P}=1.20\,\text{A} \] Thus, \[ I_P=6.0+1.20 \] \[ I_P=7.20\,\text{A} \] \(\textbf{Determine the paths in experiment Q:}\) The source is connected between B and C. The direct B–C path contains \(6.0\,\Omega\). The indirect B–A–C path contains \(3.0\,\Omega\) and \(9.0\,\Omega\) in series. \[ R_{\text{indirect},Q}=3.0+9.0 \] \[ R_{\text{indirect},Q}=12\,\Omega \] \(\textbf{Source current in experiment Q:}\) The direct-branch current is \[ I_{BC,Q}=\frac{18}{6.0} \] \[ I_{BC,Q}=3.0\,\text{A} \] The indirect-branch current is \[ I_{\text{indirect},Q}=\frac{18}{12} \] \[ I_{\text{indirect},Q}=1.50\,\text{A} \] Therefore, \[ I_Q=3.0+1.50 \] \[ I_Q=4.50\,\text{A} \] Hence, \[ \frac{I_P}{I_Q}=\frac{7.20}{4.50} \] \[ \frac{I_P}{I_Q}=\frac85 \] \(\textbf{Compare power in the A–B resistor:}\) In experiment P, the A–B resistor is connected directly across the source. \[ P_{AB,P}=\frac{18^2}{3.0} \] \[ P_{AB,P}=108\,\text{W} \] In experiment Q, the same resistor lies in the indirect B–A–C path and carries \(1.50\,\text{A}\). \[ P_{AB,Q}=(1.50)^2(3.0) \] \[ P_{AB,Q}=6.75\,\text{W} \] Therefore, \[ \frac{P_{AB,P}}{P_{AB,Q}} =\frac{108}{6.75} \] \[ \frac{P_{AB,P}}{P_{AB,Q}}=16 \] The resistors and source voltage are unchanged, but selecting a different pair of source terminals changes the two available parallel paths. The A–B resistor changes from being directly across the source to being part of a longer series path. Its current and power consequently change substantially. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_P}{I_Q}=\frac85,\quad\frac{P_{AB,P}}{P_{AB,Q}}=16} \]

Explanation: \(\textbf{Interpret the three original conducting paths:}\) Although the circuit contains four resistors, there are three parallel paths between source terminals A and B. The upper path contains two resistors in series and therefore has resistance \(2R\). Each of the two lower paths contains one resistor of resistance \(R\). \(\textbf{Initial equivalent resistance:}\) Parallel conductances add: \[ \frac1{R_i}=\frac1{2R}+\frac1R+\frac1R \] \[ \frac1{R_i}=\frac{5}{2R} \] Therefore, \[ R_i=\frac{2R}{5} \] \(\textbf{Infer the individual resistor resistance:}\) The initial source voltage is \[ V=12\,\text{V} \] The measured source current is \[ I_i=10\,\text{A} \] Therefore, \[ R_i=\frac{12}{10} \] \[ R_i=1.20\,\Omega \] Equating the two expressions, \[ \frac{2R}{5}=1.20 \] \[ R=3.0\,\Omega \] \(\textbf{Equivalent resistance after disconnection:}\) One lower branch is removed. The remaining paths have resistances \(2R\) and \(R\). Therefore, \[ \frac1{R_f}=\frac1{2R}+\frac1R \] \[ \frac1{R_f}=\frac3{2R} \] \[ R_f=\frac{2R}{3} \] Substituting \(R=3.0\,\Omega\), \[ R_f=2.0\,\Omega \] \(\textbf{Final source current:}\) \[ I_f=\frac{12}{2.0} \] \[ I_f=6.0\,\text{A} \] \(\textbf{Power in an upper-path resistor:}\) The upper path remains directly connected across the same ideal source in both arrangements. Its resistance is \[ R_{\text{upper}}=2R=6.0\,\Omega \] Therefore, its current remains \[ I_{\text{upper}}=\frac{12}{6.0} \] \[ I_{\text{upper}}=2.0\,\text{A} \] Each resistor in that path has resistance \(3.0\,\Omega\). Hence, \[ P_{\text{upper resistor}}=I_{\text{upper}}^2R \] \[ P_{\text{upper resistor}}=(2.0)^2(3.0) \] \[ P_{\text{upper resistor}}=12\,\text{W} \] Disconnecting a lower parallel branch reduces the total source current. However, the upper path remains connected across the same fixed voltage and therefore retains its original current and power. The source-current change must not be confused with a change in every branch current. \(\textbf{Final answer:}\) \[ \boxed{I_f=6.0\,\text{A},\quad P_{\text{upper resistor}}=12\,\text{W}} \]

Explanation: \(\textbf{Original divider current:}\) Initially, P and Q are connected in series. \[ R_i=4.0+12 \] \[ R_i=16\,\Omega \] Therefore, \[ I_i=\frac{12}{16} \] \[ I_i=0.75\,\text{A} \] \(\textbf{Original voltage and power in Q:}\) \[ V_{Q,i}=I_iR_Q \] \[ V_{Q,i}=(0.75)(12) \] \[ V_{Q,i}=9.0\,\text{V} \] The original power in Q is \[ P_{Q,i}=\frac{9.0^2}{12} \] \[ P_{Q,i}=6.75\,\text{W} \] \(\textbf{Use the unchanged-power requirement:}\) The \(12\,\Omega\) resistor remains unchanged. To maintain the same power in it, \[ P_Q=\frac{V_Q^2}{R_Q} \] must retain its original value. With positive source polarity, the required voltage across Q therefore remains \[ V_{Q,f}=9.0\,\text{V} \] \(\textbf{Determine the current in the added branch:}\) Resistor X is connected in parallel with Q. Both experience \(9.0\,\text{V}\). Thus, \[ I_X=\frac{9.0}{6.0} \] \[ I_X=1.50\,\text{A} \] \(\textbf{Current through the original Q branch:}\) \[ I_Q=\frac{9.0}{12} \] \[ I_Q=0.75\,\text{A} \] \(\textbf{Current through P after loading:}\) P carries the sum of the two parallel-branch currents. \[ I_{P,f}=I_Q+I_X \] \[ I_{P,f}=0.75+1.50 \] \[ I_{P,f}=2.25\,\text{A} \] \(\textbf{Required new source voltage:}\) The voltage across P is \[ V_{P,f}=I_{P,f}R_P \] \[ V_{P,f}=(2.25)(4.0) \] \[ V_{P,f}=9.0\,\text{V} \] The complete source voltage must therefore be \[ V_f=V_{P,f}+V_{Q,f} \] \[ V_f=9.0+9.0 \] \[ V_f=18\,\text{V} \] \(\textbf{Compare powers in P:}\) Since P has constant resistance, \[ \frac{P_{P,f}}{P_{P,i}} =\left(\frac{I_{P,f}}{I_i}\right)^2 \] Substituting, \[ \frac{P_{P,f}}{P_{P,i}} =\left(\frac{2.25}{0.75}\right)^2 \] \[ \frac{P_{P,f}}{P_{P,i}}=9 \] The voltage across Q is deliberately maintained at its original value to preserve its power. However, adding X creates another current path, so the current through the common series resistor P increases substantially. Keeping one resistor's power unchanged can therefore require much greater power dissipation elsewhere in the circuit. \(\textbf{Final answer:}\) \[ \boxed{V_f=18\,\text{V},\quad\frac{P_{P,f}}{P_{P,i}}=9} \]

Explanation: Statement I is correct because P and Q initially experience the same potential difference. For parallel resistors, current is inversely proportional to resistance. Since P has resistance \(4R\) while Q has resistance \(R\), Q carries four times the current of P. Statement II is correct because the additional resistor is connected in series with the original parallel combination. Under steady current, a positive potential difference develops across this added resistor. The voltage remaining across the parallel combination is therefore less than the full source voltage. Statement III is incorrect because P and Q still share the same two nodes after the additional resistor is connected. Their individual resistances remain unchanged. Both branch currents decrease because their common applied voltage decreases, but their ratio remains fixed by their resistances. Thus, the ratio \(I_Q/I_P\) remains four. Statement IV is correct because a positive resistance has been added in series with the original network. This increases the equivalent resistance of the complete circuit. The ideal source maintains the same voltage, so the total current drawn from it decreases. The important distinction is between the absolute currents in the two parallel branches and their ratio. The added series resistance changes the voltage supplied to the parallel section, thereby changing both branch currents together. It does not alter the inverse-resistance current-division relationship between those unchanged branches. Therefore, Statements I, II and IV are correct.

Explanation: \(\textbf{Relate total power to equivalent resistance:}\) Each arrangement is supplied by the same ideal voltage source. Therefore, \[ P=\frac{V^2}{R_{\text{eq}}} \] A smaller equivalent resistance draws greater total power at the same source voltage. The three equivalent resistances must first be determined from their actual connections. \(\textbf{Arrangement P — All resistors in series:}\) \[ R_P=4.0+6.0+12 \] \[ R_P=22\,\Omega \] \(\textbf{Arrangement Q — Series-parallel combination:}\) The parallel section has resistance \[ R_{6\parallel12}=\frac{(6.0)(12)}{6.0+12} \] \[ R_{6\parallel12}=4.0\,\Omega \] The additional series resistor gives \[ R_Q=4.0+4.0 \] \[ R_Q=8.0\,\Omega \] \(\textbf{Arrangement R — Parallel paths:}\) The \(4.0\,\Omega\) and \(6.0\,\Omega\) resistors are in series. \[ R_{4+6}=4.0+6.0 \] \[ R_{4+6}=10\,\Omega \] This path is in parallel with \(12\,\Omega\). Therefore, \[ R_R=\frac{(10)(12)}{10+12} \] \[ R_R=\frac{60}{11}\,\Omega \] \(\textbf{Order the equivalent resistances:}\) The resistance values satisfy \[ 22\gt 8\gt \frac{60}{11} \] Therefore, under the same source voltage, the power order is reversed: \[ P_P\lt P_Q\lt P_R \] \(\textbf{Determine the requested power ratio:}\) \[ \frac{P_R}{P_Q} =\frac{V^2/R_R}{V^2/R_Q} \] \[ \frac{P_R}{P_Q}=\frac{R_Q}{R_R} \] Substituting, \[ \frac{P_R}{P_Q}=\frac8{60/11} \] \[ \frac{P_R}{P_Q}=\frac{88}{60} \] \[ \frac{P_R}{P_Q}=\frac{22}{15} \] Rearranging the same three resistors can change the total electrical power substantially without changing any individual resistor value. The power ranking must be based on complete equivalent resistances rather than on the number of resistors appearing in parallel. \(\textbf{Final answer:}\) The increasing order of total power is \[ P_P\lt P_Q\lt P_R \] The corresponding power ratio is \[ \frac{P_R}{P_Q}=\frac{22}{15} \]

Explanation: \(\textbf{Initial branch resistances:}\) The upper branch contains P and Q in series: \[ R_{\text{upper},i}=2.0+4.0 \] \[ R_{\text{upper},i}=6.0\,\Omega \] The lower branch contains R and S: \[ R_{\text{lower},i}=2.0+4.0 \] \[ R_{\text{lower},i}=6.0\,\Omega \] The two equal-resistance branches are connected in parallel. \(\textbf{Initial source current:}\) Their equivalent resistance is \[ R_i=\frac{(6.0)(6.0)}{6.0+6.0} \] \[ R_i=3.0\,\Omega \] Therefore, \[ I_i=\frac{12}{3.0} \] \[ I_i=4.0\,\text{A} \] Each identical branch carries \(2.0\,\text{A}\). \(\textbf{Initial power in resistor Q:}\) Resistor Q belongs to the upper branch. \[ P_{Q,i}=I_{\text{upper},i}^2R_Q \] \[ P_{Q,i}=(2.0)^2(4.0) \] \[ P_{Q,i}=16\,\text{W} \] \(\textbf{Branch resistances after exchange:}\) The new upper branch contains P and R: \[ R_{\text{upper},f}=2.0+2.0 \] \[ R_{\text{upper},f}=4.0\,\Omega \] The new lower branch contains Q and S: \[ R_{\text{lower},f}=4.0+4.0 \] \[ R_{\text{lower},f}=8.0\,\Omega \] \(\textbf{Determine the final source current:}\) The two new branches remain in parallel. Their currents are \[ I_{\text{upper},f}=\frac{12}{4.0}=3.0\,\text{A} \] \[ I_{\text{lower},f}=\frac{12}{8.0}=1.50\,\text{A} \] Therefore, \[ I_f=3.0+1.50 \] \[ I_f=4.50\,\text{A} \] \(\textbf{Final power in resistor Q:}\) Q now belongs to the lower branch and carries \(1.50\,\text{A}\). \[ P_{Q,f}=(1.50)^2(4.0) \] \[ P_{Q,f}=9.0\,\text{W} \] Thus, \[ \frac{P_{Q,f}}{P_{Q,i}}=\frac{9.0}{16} \] \[ \frac{P_{Q,f}}{P_{Q,i}}=\frac9{16} \] The total source current increases after the exchange because the new parallel arrangement has a smaller equivalent resistance. However, resistor Q moves into the higher-resistance branch, so its own current and power decrease. This illustrates why total circuit behaviour and individual resistor behaviour must be analysed separately. \(\textbf{Final answer:}\) \[ \boxed{I_f=4.50\,\text{A},\quad\frac{P_{Q,f}}{P_{Q,i}}=\frac9{16}} \]

Explanation: \(\textbf{Determine the parallel-section resistance:}\) The parallel section contains \(3.0\,\Omega\) and \(6.0\,\Omega\). \[ R_p=\frac{(3.0)(6.0)}{3.0+6.0} \] \[ R_p=2.0\,\Omega \] \(\textbf{Determine the complete circuit resistance:}\) The separate \(6.0\,\Omega\) resistor is in series with the parallel section. \[ R_{\text{total}}=6.0+2.0 \] \[ R_{\text{total}}=8.0\,\Omega \] \(\textbf{Apply the power limit of the series resistor:}\) The series resistor carries the complete source current. Its maximum permissible power is \[ P_{s,\max}=12\,\text{W} \] Using \[ P_s=I^2R_s, \] the limiting current is \[ I_{s,\max}=\sqrt{\frac{12}{6.0}} \] \[ I_{s,\max}=\sqrt2\,\text{A} \] This would correspond to a source voltage \[ V_{s,\max}=I_{s,\max}R_{\text{total}} \] \[ V_{s,\max}=8\sqrt2\,\text{V} \] \(\textbf{Apply the parallel-branch power limit:}\) Let the source current be \(I\). The potential difference across the parallel section is \[ V_p=IR_p \] \[ V_p=2I \] The power in the \(3.0\,\Omega\) branch resistor is \[ P_3=\frac{V_p^2}{3.0} \] Substituting \(V_p=2I\), \[ P_3=\frac{4I^2}{3} \] The allowed maximum is \(3.0\,\text{W}\). Therefore, \[ \frac{4I^2}{3}\leq3 \] \[ I^2\leq\frac94 \] \[ I\leq1.50\,\text{A} \] This branch permits a larger current than the series resistor because \[ 1.50\gt \sqrt2 \] Therefore, the series resistor imposes the actual safety limit. \(\textbf{Greatest safe source voltage:}\) The maximum safe current is \(\sqrt2\,\text{A}\). Thus, \[ V_{\max}=I_{\max}R_{\text{total}} \] \[ V_{\max}=8\sqrt2\,\text{V} \] \(\textbf{Power in the }3.0\,\Omega\textbf{ branch at this limit:}\) \[ P_3=\frac{4I_{\max}^2}{3} \] \[ P_3=\frac{4(2)}{3} \] \[ P_3=\frac83\,\text{W} \] This is below the branch's \(3.0\,\text{W}\) limit. The actual maximum source voltage is set by the more restrictive of the two independent component-power constraints. A resistor with the smaller nominal power rating is not necessarily the first component to reach its limit, because branch currents differ. \(\textbf{Final answer:}\) \[ \boxed{V_{\max}=8\sqrt2\,\text{V},\quad P_3=\frac83\,\text{W}} \]

Explanation: \(\textbf{Infer the original equivalent resistance:}\) The source voltage is \[ V=12\,\text{V} \] The initial source current is \[ I_i=2.0\,\text{A} \] Therefore, \[ R_{\text{eq},i}=\frac{12}{2.0} \] \[ R_{\text{eq},i}=6.0\,\Omega \] \(\textbf{Calculate the original parallel resistance:}\) The parallel section contains \(4.0\,\Omega\) and \(12\,\Omega\). \[ R_p=\frac{(4.0)(12)}{4.0+12} \] \[ R_p=3.0\,\Omega \] \(\textbf{Determine the unknown series resistance:}\) \[ R_{\text{eq},i}=R+R_p \] Thus, \[ 6.0=R+3.0 \] \[ R=3.0\,\Omega \] \(\textbf{Initial voltage across the parallel section:}\) \[ V_{p,i}=I_iR_p \] \[ V_{p,i}=(2.0)(3.0) \] \[ V_{p,i}=6.0\,\text{V} \] \(\textbf{Initial power in the }4.0\,\Omega\textbf{ resistor:}\) \[ P_{4,i}=\frac{V_{p,i}^2}{4.0} \] \[ P_{4,i}=\frac{6.0^2}{4.0} \] \[ P_{4,i}=9.0\,\text{W} \] \(\textbf{Changed network after disconnecting one branch:}\) The \(12\,\Omega\) branch is open. Only \(R=3.0\,\Omega\) and the \(4.0\,\Omega\) resistor remain in series. Thus, \[ R_{\text{eq},f}=3.0+4.0 \] \[ R_{\text{eq},f}=7.0\,\Omega \] The new source current is \[ I_f=\frac{12}{7.0} \] \[ I_f=\frac{12}{7}\,\text{A} \] \(\textbf{Final power in the surviving resistor:}\) The surviving \(4.0\,\Omega\) resistor carries the complete final series current. \[ P_{4,f}=I_f^2(4.0) \] \[ P_{4,f}=\left(\frac{12}{7}\right)^2(4.0) \] \[ P_{4,f}=\frac{576}{49}\,\text{W} \] \(\textbf{Compare the powers:}\) \[ \frac{P_{4,f}}{P_{4,i}}=\frac{576/49}{9.0} \] \[ \frac{P_{4,f}}{P_{4,i}}=\frac{64}{49} \] The total source current decreases after removing a parallel path. However, the current through the surviving \(4.0\,\Omega\) resistor increases from \(1.50\,\text{A}\) to \(12/7\,\text{A}\). Its power consequently increases despite the lower source current. The correct change depends on the current through the individual resistor rather than on the total current alone. \(\textbf{Final answer:}\) \[ \boxed{R=3.0\,\Omega,\quad\frac{P_{4,f}}{P_{4,i}}=\frac{64}{49}} \]

Explanation: \(\textbf{Resistance of the initially connected branch:}\) The upper branch contains \(3.0\,\Omega\) and \(6.0\,\Omega\) in series. \[ R_{\text{upper}}=3.0+6.0 \] \[ R_{\text{upper}}=9.0\,\Omega \] Initially, this is the only conducting path between the source terminals. \(\textbf{Resistance of the added branch:}\) The lower branch contains \(4.0\,\Omega\) and \(12\,\Omega\) in series. Therefore, \[ R_{\text{lower}}=4.0+12 \] \[ R_{\text{lower}}=16\,\Omega \] \(\textbf{Interpret the measured current increase:}\) The source is ideal and maintains the same voltage after the switch closes. The current in the previously connected upper branch therefore remains unchanged. The entire increase in source current is supplied to the newly connected lower branch. Hence, \[ I_{\text{lower}}=\Delta I_{\text{source}} \] \[ I_{\text{lower}}=1.50\,\text{A} \] \(\textbf{Determine the source voltage:}\) The lower branch experiences the full source voltage. \[ V=I_{\text{lower}}R_{\text{lower}} \] \[ V=(1.50)(16) \] \[ V=24\,\text{V} \] \(\textbf{Current in the upper branch:}\) The upper branch experiences the same \(24\,\text{V}\) before and after switching. \[ I_{\text{upper}}=\frac{24}{9.0} \] \[ I_{\text{upper}}=\frac83\,\text{A} \] This current passes through both series resistors in that branch. \(\textbf{Compare power in the }6.0\,\Omega\textbf{ resistor:}\) Its electrical power is \[ P_6=I_{\text{upper}}^2(6.0) \] Neither the branch current nor the resistor value changes after the switch closes. Therefore, \[ P_{6,f}=P_{6,i} \] Hence, \[ \frac{P_{6,f}}{P_{6,i}}=1 \] \(\textbf{Physical interpretation:}\) The lower branch draws additional current from the ideal source. However, the original upper branch remains connected across the same voltage. Its individual resistor currents and powers are consequently unaffected. The measured increase in source current directly reveals the current through the newly connected branch. That added-branch current and resistance together determine the previously unknown source voltage. \(\textbf{Final answer:}\) \[ \boxed{V=24\,\text{V},\quad\frac{P_{6,f}}{P_{6,i}}=1} \]

Explanation: \(\textbf{Begin with the innermost parallel pair:}\) The lower B–C path contains a \(6.0\,\Omega\) resistor in parallel with a \(3.0\,\Omega\) resistor. Their equivalent resistance is \[ R_{\text{inner}}=\frac{(6.0)(3.0)}{6.0+3.0} \] \[ R_{\text{inner}}=2.0\,\Omega \] \(\textbf{Complete the lower path:}\) The innermost parallel combination is in series with the \(4.0\,\Omega\) resistor. \[ R_{\text{lower}}=4.0+2.0 \] \[ R_{\text{lower}}=6.0\,\Omega \] \(\textbf{Combine the two B–C paths:}\) The upper path has resistance \(6.0\,\Omega\). The lower path also has resistance \(6.0\,\Omega\). Thus, the two B–C paths are equal-resistance parallel branches. \[ R_{BC}=\frac{(6.0)(6.0)}{6.0+6.0} \] \[ R_{BC}=3.0\,\Omega \] \(\textbf{Determine total network resistance:}\) The \(2.0\,\Omega\) resistor between A and B is in series with the B–C combination. \[ R_{\text{total}}=2.0+3.0 \] \[ R_{\text{total}}=5.0\,\Omega \] \(\textbf{Determine source current:}\) \[ I_{\text{source}}=\frac{20}{5.0} \] \[ I_{\text{source}}=4.0\,\text{A} \] \(\textbf{Voltage across the B–C network:}\) \[ V_{BC}=I_{\text{source}}R_{BC} \] \[ V_{BC}=(4.0)(3.0) \] \[ V_{BC}=12\,\text{V} \] \(\textbf{Current in the upper B–C path:}\) \[ I_{\text{upper}}=\frac{12}{6.0} \] \[ I_{\text{upper}}=2.0\,\text{A} \] The lower path also carries \(2.0\,\text{A}\). \(\textbf{Voltage across the innermost pair:}\) The lower-path current passes through its \(4.0\,\Omega\) series resistor. The voltage across that resistor is \[ V_4=(2.0)(4.0) \] \[ V_4=8.0\,\text{V} \] Therefore, the voltage across the innermost parallel pair is \[ V_{\text{inner}}=12-8.0 \] \[ V_{\text{inner}}=4.0\,\text{V} \] \(\textbf{Current through the }3.0\,\Omega\textbf{ resistor:}\) \[ I_3=\frac{V_{\text{inner}}}{3.0} \] \[ I_3=\frac{4.0}{3.0} \] \[ I_3=\frac43\,\text{A} \] The remaining \(2/3\,\text{A}\) flows through the innermost \(6.0\,\Omega\) resistor. Together, these currents account for the complete \(2.0\,\text{A}\) lower-path current. \(\textbf{Power in the upper }6.0\,\Omega\textbf{ resistor:}\) The upper resistor carries \(2.0\,\text{A}\). \[ P_{\text{upper}}=I_{\text{upper}}^2(6.0) \] \[ P_{\text{upper}}=(2.0)^2(6.0) \] \[ P_{\text{upper}}=24\,\text{W} \] The circuit contains parallel sections nested inside a larger parallel arrangement. Reduction must begin with the innermost pair, after which the branch currents and internal voltages can be recovered in reverse order. Applying the total source current to the innermost resistor would violate the junction-current conditions. \(\textbf{Final answer:}\) \[ \boxed{I_3=\frac43\,\text{A},\quad P_{\text{upper}}=24\,\text{W}} \]

Explanation: \(\textbf{Assign conventional-current directions:}\) A positive ion moving right produces conventional current towards the right. A negative ion moving left also produces conventional current towards the right. However, a negative ion moving right contributes conventional current towards the left. Let \(I_+\) and \(I_-\) represent the positive magnitudes of the individual ionic-current contributions. \(\textbf{Experiment P — Contributions add:}\) Both species contribute conventional current towards the right. Therefore, \[ I_++I_-=0.176\,\text{A} \] \(\textbf{Experiment Q — Contributions oppose:}\) The negative ions reverse their velocity, while the positive-ion motion remains unchanged. Therefore, \[ I_+-I_-=0.016\,\text{A} \] \(\textbf{Recover the individual currents:}\) Adding the two equations, \[ 2I_+=0.176+0.016 \] \[ I_+=0.096\,\text{A} \] Subtracting, \[ 2I_-=0.176-0.016 \] \[ I_-=0.080\,\text{A} \] \(\textbf{Positive-ion crossing rate:}\) Each positive ion carries charge \(+2e\). If \(N_+\) positive ions cross per second, \[ I_+=2eN_+ \] Therefore, \[ N_+=\frac{0.096}{2(1.6\times10^{-19})} \] \[ N_+=3.0\times10^{17}\,\text{s}^{-1} \] \(\textbf{Negative-ion share in experiment P:}\) The negative-ion current contribution is \(0.080\,\text{A}\), while the total is \(0.176\,\text{A}\). Thus, \[ f_-=\frac{0.080}{0.176} \] \[ f_-=\frac5{11} \] The measured current is an algebraic sum of charge-transport contributions, not simply the total number of ions crossing. Reversing the motion of a negative carrier reverses its contribution to conventional current. The two experiments provide independent equations, allowing the individual ionic contributions to be determined without initially knowing either crossing rate. \(\textbf{Final answer:}\) \[ \boxed{N_+=3.0\times10^{17}\,\text{s}^{-1},\quad f_-=\frac5{11}} \]

Explanation: \(\textbf{Recover the components of current density:}\) Current crossing a plane section is determined by the component of current density along its normal. For uniform current density, \[ I=\vec j\cdot\vec A \] The section normals are along mutually perpendicular coordinate axes. Thus, the first measurement determines \(j_x\), and the second determines \(j_y\). \(\textbf{Current-density components:}\) Convert the common area: \[ A=1.0\,\text{mm}^2=1.0\times10^{-6}\,\text{m}^2 \] Therefore, \[ j_x=\frac{4.0}{1.0\times10^{-6}} \] \[ j_x=4.0\times10^6\,\text{A m}^{-2} \] Similarly, \[ j_y=\frac{3.0}{1.0\times10^{-6}} \] \[ j_y=3.0\times10^6\,\text{A m}^{-2} \] Both components are positive. \(\textbf{Determine the magnitude of current density:}\) The components are perpendicular, so \[ |\vec j|=\sqrt{j_x^2+j_y^2} \] \[ |\vec j|=\sqrt{4^2+3^2}\times10^6 \] \[ |\vec j|=5.0\times10^6\,\text{A m}^{-2} \] \(\textbf{Convert current density into electron drift speed:}\) For electron conduction, \[ |\vec j|=ne|\vec v_d| \] Therefore, \[ |\vec v_d|=\frac{|\vec j|}{ne} \] Substituting, \[ |\vec v_d| =\frac{5.0\times10^6}{(5.0\times10^{28})(1.6\times10^{-19})} \] \[ |\vec v_d|=6.25\times10^{-4}\,\text{m s}^{-1} \] \[ |\vec v_d|=0.625\,\text{mm s}^{-1} \] \(\textbf{Determine the drift direction:}\) Conventional current density points into the first quadrant because its \(x\)- and \(y\)-components are positive. Electrons carry negative charge. Their average drift velocity is opposite to conventional current density, placing it in the third quadrant. The two current measurements must be treated as perpendicular vector components, not added as scalar currents through one cross-section. \(\textbf{Final answer:}\) \[ \boxed{|\vec v_d|=0.625\,\text{mm s}^{-1},\quad\text{third quadrant}} \]