Current Electricity MCQs with Answers – Part 4 (NEET Physics)

Explanation: \(\textbf{Determine the initial branch currents:}\) The elements are connected in parallel, so both experience the full \(12\,\text{V}\). Initially, \[ I_{M,0}=\frac{12}{6.0} \] \[ I_{M,0}=2.0\,\text{A} \] For N, \[ I_{N,0}=\frac{12}{12} \] \[ I_{N,0}=1.0\,\text{A} \] Therefore, \[ I_{\text{source},0}=2.0+1.0 \] \[ I_{\text{source},0}=3.0\,\text{A} \] \(\textbf{Resistance of M after heating:}\) For the specified linear model, \[ R_T=R_0(1+\alpha_0T) \] Thus, \[ R_{M,100}=6.0[1+(0.005)(100)] \] \[ R_{M,100}=6.0(1.5) \] \[ R_{M,100}=9.0\,\Omega \] \(\textbf{Resistance of N after heating:}\) \[ R_{N,100}=12[1-(0.004)(100)] \] \[ R_{N,100}=12(0.6) \] \[ R_{N,100}=7.2\,\Omega \] \(\textbf{Recalculate the branch currents:}\) The ideal source continues to maintain \(12\,\text{V}\). For M, \[ I_{M,100}=\frac{12}{9.0} \] \[ I_{M,100}=\frac43\,\text{A} \] For N, \[ I_{N,100}=\frac{12}{7.2} \] \[ I_{N,100}=\frac53\,\text{A} \] \(\textbf{Compare total source currents:}\) \[ I_{\text{source},100}=\frac43+\frac53 \] \[ I_{\text{source},100}=3.0\,\text{A} \] Hence, \[ \frac{I_{\text{source},100}}{I_{\text{source},0}} =\frac{3.0}{3.0} \] \[ \frac{I_{\text{source},100}}{I_{\text{source},0}}=1 \] \(\textbf{Compare currents through N:}\) \[ \frac{I_{N,100}}{I_{N,0}} =\frac{5/3}{1} \] \[ \frac{I_{N,100}}{I_{N,0}}=\frac53 \] \(\textbf{Physical interpretation:}\) The current through M decreases because its resistance increases. The current through N increases because its resistance decreases. For these particular values, the changes exactly compensate in the total source current. The equivalent resistance of the complete parallel network therefore has the same value at the two specified temperatures, even though neither element retains its original resistance. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{\text{source},100}}{I_{\text{source},0}}=1,\quad\frac{I_{N,100}}{I_{N,0}}=\frac53} \]

Explanation: \(\textbf{Identify the steady thermal condition:}\) The element receives electrical energy from the source and loses thermal energy to the surroundings. At a steady temperature, it does not accumulate thermal energy. Therefore, \[ P_{\text{electrical}}=P_{\text{loss}} \] The electrical heating rate depends on temperature because the element's resistance changes. \(\textbf{Express the temperature-dependent resistance:}\) Let \[ x=T-20 \] where \(x\) is the temperature rise in kelvin. The resistance becomes \[ R=6.0(1+0.010x)\,\Omega \] \(\textbf{Express electrical power:}\) The source voltage remains \(12\,\text{V}\). For an ohmic element at its maintained temperature, \[ P_{\text{electrical}}=\frac{V^2}{R} \] Therefore, \[ P_{\text{electrical}}=\frac{12^2}{6.0(1+0.010x)} \] \[ P_{\text{electrical}}=\frac{24}{1+0.010x}\,\text{W} \] \(\textbf{Express the thermal-loss rate:}\) The stipulated loss rate is \[ P_{\text{loss}}=(1.0\,\text{W K}^{-1})x \] Thus, its numerical value in watts is \(x\). \(\textbf{Construct the steady-state equation:}\) \[ \frac{24}{1+0.010x}=x \] Multiplying through, \[ 24=x(1+0.010x) \] \[ 24=x+0.010x^2 \] Multiplying by \(100\), \[ x^2+100x-2400=0 \] Factorising, \[ (x-20)(x+120)=0 \] The solutions are \(x=20\) and \(x=-120\). The negative solution would place the element below the stated reference range and is not compatible with the heating condition. Therefore, \[ x=20\,\text{K} \] \(\textbf{Determine the steady temperature:}\) \[ T=20+20 \] \[ T=40^\circ\text{C} \] \(\textbf{Find the corresponding current:}\) At \(40^\circ\text{C}\), \[ R=6.0[1+0.010(20)] \] \[ R=7.2\,\Omega \] Thus, \[ I=\frac{12}{7.2} \] \[ I=\frac53\,\text{A} \] \(\textbf{Check the energy balance:}\) The electrical power is \[ P=(12)\left(\frac53\right)=20\,\text{W} \] The thermal-loss rate at a \(20\,\text{K}\) temperature rise is also \(20\,\text{W}\). The equality confirms the steady condition. The source voltage cannot be used with the cold resistance alone because the final resistance depends on the temperature that must first be determined. \(\textbf{Final answer:}\) \[ \boxed{T=40^\circ\text{C},\quad I=\frac53\,\text{A}} \]

Explanation: \(\textbf{Use the fixed material volume:}\) The complete material forms one uniform wire. Its volume is therefore \[ \mathcal V=AL \] Convert the given volume: \[ \mathcal V=8.0\,\text{cm}^3 \] \[ \mathcal V=8.0\times10^{-6}\,\text{m}^3 \] \(\textbf{Eliminate the unknown wire length:}\) For a uniform conducting wire, \[ R=\frac{\rho L}{A} \] From the volume condition, \[ L=\frac{\mathcal V}{A} \] Substituting, \[ R=\frac{\rho\mathcal V}{A^2} \] This relation is appropriate because the material volume, rather than the wire length, is specified independently. \(\textbf{Determine the required area:}\) Rearranging, \[ A^2=\frac{\rho\mathcal V}{R} \] Substituting the given values, \[ A^2=\frac{(2.0\times10^{-6})(8.0\times10^{-6})}{4.0} \] \[ A^2=4.0\times10^{-12}\,\text{m}^4 \] Taking the positive square root, \[ A=2.0\times10^{-6}\,\text{m}^2 \] \[ A=2.0\,\text{mm}^2 \] \(\textbf{Check the corresponding wire length:}\) \[ L=\frac{\mathcal V}{A} \] \[ L=\frac{8.0\times10^{-6}}{2.0\times10^{-6}} \] \[ L=4.0\,\text{m} \] Using the usual resistance relation gives \[ R=\frac{(2.0\times10^{-6})(4.0)}{2.0\times10^{-6}} \] \[ R=4.0\,\Omega \] \(\textbf{Current under the specified source voltage:}\) \[ I=\frac VR \] \[ I=\frac{12}{4.0} \] \[ I=3.0\,\text{A} \] \(\textbf{Determine current density:}\) \[ j=\frac IA \] \[ j=\frac{3.0}{2.0\times10^{-6}} \] \[ j=1.5\times10^6\,\text{A m}^{-2} \] With fixed material volume, increasing cross-sectional area necessarily shortens the wire. Both geometrical changes reduce resistance, producing an inverse-square dependence on area. The electrical design requirement therefore uniquely determines the cross-sectional area before current density can be calculated. \(\textbf{Final answer:}\) \[ \boxed{A=2.0\,\text{mm}^2,\quad j=1.5\times10^6\,\text{A m}^{-2}} \]

Explanation: \(\textbf{Use the parallel conduction condition:}\) The two materials share the same end terminals and conductor length. Their electric-field magnitudes are therefore equal. For an isotropic ohmic material, \[ j=\sigma E \] Consequently, the current in each region is \[ I=\sigma AE \] \(\textbf{Construct the complete current relation:}\) The total current is the sum of the currents through P and Q: \[ I_{\text{total}}=I_P+I_Q \] Thus, \[ I_{\text{total}}=\sigma_PA_PE+\sigma_QA_QE \] Using the effective-conductivity definition, \[ \sigma_{\text{eff}}A_{\text{total}}E =\sigma_PA_PE+\sigma_QA_QE \] Cancelling the common electric-field magnitude, \[ \sigma_{\text{eff}}A_{\text{total}} =\sigma_PA_P+\sigma_QA_Q \] \(\textbf{Apply the total-area constraint:}\) \[ A_P+A_Q=A_{\text{total}} \] Therefore, \[ A_P=A_{\text{total}}-A_Q \] Substituting, \[ \sigma_{\text{eff}}A_{\text{total}} =\sigma_P(A_{\text{total}}-A_Q)+\sigma_QA_Q \] Rearranging, \[ (\sigma_{\text{eff}}-\sigma_P)A_{\text{total}} =(\sigma_Q-\sigma_P)A_Q \] \(\textbf{Determine Q's area fraction:}\) \[ \frac{A_Q}{A_{\text{total}}} =\frac{\sigma_{\text{eff}}-\sigma_P}{\sigma_Q-\sigma_P} \] Substituting conductivities in units of \(10^6\,\text{S m}^{-1}\), \[ \frac{A_Q}{A_{\text{total}}}=\frac{2-1}{4-1} \] \[ \frac{A_Q}{A_{\text{total}}}=\frac13 \] Hence, \[ A_Q=\frac13(3.0) \] \[ A_Q=1.0\,\text{mm}^2 \] \(\textbf{Determine Q's share of total current:}\) Because the field is common, \[ \frac{I_Q}{I_{\text{total}}} =\frac{\sigma_QA_Q}{\sigma_{\text{eff}}A_{\text{total}}} \] Substituting, \[ \frac{I_Q}{I_{\text{total}}} =\frac{(4)(1.0)}{(2)(3.0)} \] \[ \frac{I_Q}{I_{\text{total}}}=\frac23 \] The material occupying only one-third of the area carries two-thirds of the total current. This occurs because Q has four times the conductivity of P. Effective conductivity is determined by the area-weighted current contributions, not by an unweighted average of the two material conductivities. \(\textbf{Final answer:}\) \[ \boxed{A_Q=1.0\,\text{mm}^2,\quad\frac{I_Q}{I_{\text{total}}}=\frac23} \]

Explanation: \(\textbf{Identify the series conduction geometry:}\) The two materials are joined end to end. They therefore carry the same steady current rather than the same voltage drop. Their cross-sectional areas are equal, so their current-density magnitudes are also equal. \(\textbf{Resistance of section P:}\) For a uniform conductor, \[ R=\frac{\rho L}{A} \] Therefore, \[ R_P=\frac{\rho_PL}{A} \] \(\textbf{Resistance of section Q:}\) Section Q has twice the length and a different resistivity. Thus, \[ R_Q=\frac{\rho_Q(2L)}{A} \] \(\textbf{Combine the series resistances:}\) \[ R_{\text{total}}=R_P+R_Q \] \[ R_{\text{total}} =\frac{\rho_PL}{A}+\frac{2\rho_QL}{A} \] Factorising, \[ R_{\text{total}} =\frac{(\rho_P+2\rho_Q)L}{A} \] \(\textbf{Determine the effective resistivity:}\) The complete rod has length \(3L\). By definition, \[ R_{\text{total}}=\frac{\rho_{\text{eff}}(3L)}{A} \] Equating the two expressions, \[ 3\rho_{\text{eff}}=\rho_P+2\rho_Q \] Therefore, \[ \rho_{\text{eff}}=\frac{\rho_P+2\rho_Q}{3} \] Substituting, \[ \rho_{\text{eff}} =\frac{2.0\times10^{-6}+2(6.0\times10^{-6})}{3} \] \[ \rho_{\text{eff}}=\frac{14}{3}\times10^{-6}\,\Omega\,\text{m} \] \(\textbf{Calculate the voltage fraction across Q:}\) The same current passes through both sections. Therefore, their voltage drops are proportional to their resistances. \[ \frac{V_Q}{V_{\text{total}}} =\frac{R_Q}{R_P+R_Q} \] Substituting the resistance expressions, \[ \frac{V_Q}{V_{\text{total}}} =\frac{2\rho_Q}{\rho_P+2\rho_Q} \] Using the given resistivities, \[ \frac{V_Q}{V_{\text{total}}} =\frac{2(6.0)}{2.0+2(6.0)} \] \[ \frac{V_Q}{V_{\text{total}}}=\frac{12}{14} \] \[ \frac{V_Q}{V_{\text{total}}}=\frac67 \] \(\textbf{Physical interpretation:}\) Section Q contributes most of the total resistance because it is both longer and more resistive. Accordingly, it receives most of the applied voltage under the common-current condition. The effective resistivity is a length-weighted average appropriate to these end-to-end sections. It must not be calculated using the parallel-conductivity relation applicable to side-by-side conducting materials. \(\textbf{Final answer:}\) \[ \boxed{\rho_{\text{eff}}=\frac{14}{3}\times10^{-6}\,\Omega\,\text{m},\quad\frac{V_Q}{V_{\text{total}}}=\frac67} \]

Explanation: \(\textbf{Extract resistance from the constant-current experiment:}\) For steady current through an ohmic resistor, \[ H=I^2Rt \] The electrical energy is \[ H=50\,\text{J} \] The maintained current is \(1.0\,\text{A}\), and the duration is \(5.0\,\text{s}\). Therefore, \[ R_T=\frac{H}{I^2t} \] Substituting, \[ R_T=\frac{50}{(1.0)^2(5.0)} \] \[ R_T=10\,\Omega \] \(\textbf{Convert resistance into temperature:}\) The linear relation referred to \(20^\circ\text{C}\) is \[ R_T=R_{20}[1+\alpha_{20}(T-20)] \] Substituting, \[ 10=8.0[1+(5.0\times10^{-3})(T-20)] \] Dividing by eight, \[ 1.25=1+0.005(T-20) \] Therefore, \[ 0.25=0.005(T-20) \] \[ T-20=50 \] \[ T=70^\circ\text{C} \] \(\textbf{Change to the constant-voltage condition:}\) In the second experiment, the source maintains \(10\,\text{V}\). For an ohmic resistor at constant voltage, \[ P=\frac{V^2}{R} \] At \(20^\circ\text{C}\), \[ P_{20}=\frac{10^2}{8.0} \] \[ P_{20}=12.5\,\text{W} \] At \(70^\circ\text{C}\), \[ P_{70}=\frac{10^2}{10} \] \[ P_{70}=10\,\text{W} \] \(\textbf{Compare electrical powers:}\) \[ \frac{P_{70}}{P_{20}}=\frac{10}{12.5} \] \[ \frac{P_{70}}{P_{20}}=\frac45 \] \(\textbf{Interpret the change of source constraint:}\) The first experiment uses a fixed current, so greater resistance would produce greater heating power. The second experiment uses a fixed voltage, so greater resistance produces smaller heating power. The same temperature-dependent resistor can therefore show opposite power trends under the two source constraints. \(\textbf{Final answer:}\) \[ \boxed{T=70^\circ\text{C},\quad\frac{P_{70}}{P_{20}}=\frac45} \]

Explanation: \(\textbf{Convert conductivity change into resistivity change:}\) Conductivity and resistivity are reciprocal: \[ \sigma=\frac1\rho \] Therefore, \[ \frac{\rho_{120}}{\rho_{20}} =\frac{\sigma_{20}}{\sigma_{120}} \] Substituting, \[ \frac{\rho_{120}}{\rho_{20}} =\frac{5.0\times10^6}{4.0\times10^6} \] \[ \frac{\rho_{120}}{\rho_{20}}=\frac54 \] \(\textbf{Relate the material change to specimen resistance:}\) For unchanged wire dimensions, \[ R=\frac{\rho L}{A} \] Thus, \[ \frac{R_{120}}{R_{20}} =\frac{\rho_{120}}{\rho_{20}} \] \[ \frac{R_{120}}{R_{20}}=\frac54 \] The heated resistance is consequently \[ R_{120}=8.0\left(\frac54\right) \] \[ R_{120}=10\,\Omega \] \(\textbf{Infer the temperature coefficient:}\) The temperature difference is \[ \Delta T=120-20=100\,\text{K} \] Using the stated linear relation, \[ 10=8.0(1+100\alpha_{20}) \] \[ 1.25=1+100\alpha_{20} \] Therefore, \[ \alpha_{20}=2.5\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Compare electrical powers:}\) The source voltage is the same at the two temperatures. For an ohmic resistor, \[ P=\frac{V^2}{R} \] Hence, \[ \frac{P_{120}}{P_{20}} =\frac{R_{20}}{R_{120}} \] \[ \frac{P_{120}}{P_{20}}=\frac{8.0}{10} \] \[ \frac{P_{120}}{P_{20}}=\frac45 \] The conductivity decreases by one-fifth of its original value, but the resistivity increases by one-quarter of its original value. These percentages differ because resistivity and conductivity are reciprocals. The wire's constant-voltage power follows the conductivity change when its dimensions are unchanged. \(\textbf{Final answer:}\) \[ \boxed{\alpha_{20}=2.5\times10^{-3}\,\text{K}^{-1},\quad\frac{P_{120}}{P_{20}}=\frac45} \]

Explanation: Statement I is correct because the stated resistance-temperature relation has the form of a straight-line equation. The coefficient \(\alpha_0\) and reference resistance \(R_0\) remain constant over the stipulated interval. Therefore, equal temperature increments produce equal absolute resistance increments. Statement II is incorrect because conductance is the reciprocal of resistance, not a quantity that generally differs from resistance by a fixed multiplier. Its temperature dependence is \[ G_T=\frac1{R_0[1+\alpha_0(T-T_0)]} \] This relation is not generally linear in temperature even though the denominator is linear. Statement III is correct because a fixed resistance increase produces a conductance change whose magnitude depends on the original resistance. For example, increasing resistance from \(2\,\Omega\) to \(3\,\Omega\) changes conductance by a different amount from increasing it from \(3\,\Omega\) to \(4\,\Omega\). The resistance increments are equal, but the reciprocal differences are not. Statement IV is correct because power under a fixed applied voltage is \(P=V^2/R\). Since the resistance increases with temperature, power decreases when the ideal source maintains the same voltage. The result depends on the source constraint; a fixed-current source would instead give \(P=I^2R\). The question therefore distinguishes the linear material-resistance model from the nonlinear relationships obtained by taking its reciprocal. Statements I, III and IV are correct.

Explanation: The Assertion is true for the specified linear increasing resistance-temperature characteristic. A temperature coefficient describes fractional resistance change per unit temperature change relative to a chosen reference resistance. It is not simply the absolute slope of the resistance-temperature graph. For a straight-line characteristic, the absolute slope remains unchanged when the reference temperature changes. However, the resistance itself is greater at a higher reference temperature because the graph has a positive slope. Dividing the same absolute slope by a greater reference resistance gives a smaller positive temperature coefficient. Therefore, the numerical value of a temperature coefficient can depend on the reference temperature even for one exactly straight resistance-temperature graph. The Reason states precisely this distinction between absolute slope and fractional slope. It consequently provides the correct explanation of the Assertion. The smaller coefficient does not mean that the material's absolute resistance increase per kelvin has become smaller. Under the stated linear model, that absolute increase is unchanged. The result illustrates why temperature coefficients must be interpreted together with their reference temperatures. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Distinguish equal resistance from equal resistivity:}\) Both wires have the same measured resistance. However, they have different lengths and cross-sectional areas. Equal specimen resistances do not necessarily imply equal material resistivities. \(\textbf{Determine P's resistivity:}\) For a uniform wire, \[ R=\frac{\rho L}{A} \] Therefore, \[ \rho_P=\frac{R_PA_P}{L_P} \] Substituting, \[ \rho_P=\frac{(3.0)(1.0\times10^{-6})}{0.50} \] \[ \rho_P=6.0\times10^{-6}\,\Omega\,\text{m} \] \(\textbf{Determine Q's resistivity:}\) \[ \rho_Q=\frac{R_QA_Q}{L_Q} \] \[ \rho_Q=\frac{(3.0)(2.0\times10^{-6})}{2.0} \] \[ \rho_Q=3.0\times10^{-6}\,\Omega\,\text{m} \] \(\textbf{Compare conductivities:}\) Because \[ \sigma=\frac1\rho, \] the conductivity ratio is \[ \frac{\sigma_Q}{\sigma_P}=\frac{\rho_P}{\rho_Q} \] \[ \frac{\sigma_Q}{\sigma_P}=\frac{6.0}{3.0} \] \[ \frac{\sigma_Q}{\sigma_P}=2 \] \(\textbf{Compare currents under equal terminal voltage:}\) The two wires are separately connected across the same \(6.0\,\text{V}\) source. Their resistances are both \(3.0\,\Omega\). Hence, \[ I_P=I_Q=\frac{6.0}{3.0} \] \[ I_P=I_Q=2.0\,\text{A} \] \(\textbf{Determine current-density ratio:}\) Current density in a uniform wire is \[ j=\frac IA \] Therefore, \[ \frac{j_P}{j_Q} =\frac{I_P/A_P}{I_Q/A_Q} \] Since \(I_P=I_Q\), \[ \frac{j_P}{j_Q}=\frac{A_Q}{A_P} \] \[ \frac{j_P}{j_Q}=2 \] \(\textbf{Check using electric fields:}\) At the common source voltage, \[ \frac{E_P}{E_Q}=\frac{L_Q}{L_P} \] \[ \frac{E_P}{E_Q}=4 \] Although Q is twice as conductive as P, its electric field is only one-quarter as great. Thus, \(j_P/j_Q=4/2=2\), agreeing with the current-and-area calculation. The example shows that equal total current and resistance can coexist with different resistivities, conductivities, electric fields and current densities. \(\textbf{Final answer:}\) \[ \boxed{\frac{\sigma_Q}{\sigma_P}=2,\quad\frac{j_P}{j_Q}=2} \]

Explanation: \(\textbf{Determine the initial total current:}\) At \(20^\circ\text{C}\), both elements experience \(12\,\text{V}\). For M, \[ I_{M,20}=\frac{12}{4.0} \] \[ I_{M,20}=3.0\,\text{A} \] For N, \[ I_{N,20}=\frac{12}{6.0} \] \[ I_{N,20}=2.0\,\text{A} \] Therefore, \[ I_{\text{source},20}=3.0+2.0 \] \[ I_{\text{source},20}=5.0\,\text{A} \] \(\textbf{Calculate the temperature changes:}\) The temperature rise is \[ \Delta T=120-20=100\,\text{K} \] For each element, use \[ R_T=R_{20}(1+\alpha_{20}\Delta T) \] \(\textbf{Heated resistance of M:}\) \[ R_{M,120}=4.0[1+(0.010)(100)] \] \[ R_{M,120}=4.0(2.0) \] \[ R_{M,120}=8.0\,\Omega \] \(\textbf{Heated resistance of N:}\) \[ R_{N,120}=6.0[1-(0.005)(100)] \] \[ R_{N,120}=6.0(0.50) \] \[ R_{N,120}=3.0\,\Omega \] \(\textbf{Final branch currents:}\) The ideal source voltage remains \(12\,\text{V}\). Therefore, \[ I_{M,120}=\frac{12}{8.0} \] \[ I_{M,120}=1.5\,\text{A} \] For N, \[ I_{N,120}=\frac{12}{3.0} \] \[ I_{N,120}=4.0\,\text{A} \] \(\textbf{Compare total source currents:}\) \[ I_{\text{source},120}=1.5+4.0 \] \[ I_{\text{source},120}=5.5\,\text{A} \] Thus, \[ \frac{I_{\text{source},120}}{I_{\text{source},20}} =\frac{5.5}{5.0} \] \[ \frac{I_{\text{source},120}}{I_{\text{source},20}}=\frac{11}{10} \] \(\textbf{Compare electrical powers in N:}\) For a resistor connected across a fixed voltage, \[ P=\frac{V^2}{R} \] Hence, \[ \frac{P_{N,120}}{P_{N,20}} =\frac{R_{N,20}}{R_{N,120}} \] \[ \frac{P_{N,120}}{P_{N,20}}=\frac{6.0}{3.0} \] \[ \frac{P_{N,120}}{P_{N,20}}=2 \] \(\textbf{Interpret the combined response:}\) Heating reduces the current through M because its resistance increases. The same temperature change increases the current through N because its resistance decreases. The increase in N's current is larger than the decrease in M's current, so the total source current increases. The current redistribution is governed by the different temperature dependences of the two parallel elements, not by a single equivalent temperature coefficient assigned without calculation. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{\text{source},120}}{I_{\text{source},20}}=\frac{11}{10},\quad\frac{P_{N,120}}{P_{N,20}}=2} \]

Explanation: \(\textbf{Use the continuous-wire current condition:}\) Sections P and Q are connected end to end. Therefore, the same steady current passes through both sections. For each section, \[ V=IR \] Consequently, \[ \frac{R_Q}{R_P}=\frac{V_Q}{V_P} \] Substituting the measured voltages, \[ \frac{R_Q}{R_P}=\frac{8.0}{5.0} \] \[ \frac{R_Q}{R_P}=1.6 \] \(\textbf{Apply the temperature relation:}\) The two sections have identical dimensions and material properties apart from temperature. Therefore, their resistance ratio is determined by the temperature dependence of resistivity. Using the resistance at \(0^\circ\text{C}\) as reference, \[ R_{100}=R_0(1+100\alpha_0) \] Hence, \[ 1.6=1+100\alpha_0 \] \[ 100\alpha_0=0.6 \] \[ \alpha_0=6.0\times10^{-3}\,\text{K}^{-1} \] \(\textbf{Determine the power fraction:}\) Each section dissipates power \[ P=I^2R \] The current is common, so the power ratio equals the resistance ratio. Therefore, \[ \frac{P_Q}{P_P+P_Q} =\frac{R_Q}{R_P+R_Q} \] Since \(R_Q/R_P=8/5\), \[ \frac{P_Q}{P_{\text{total}}} =\frac8{5+8} \] \[ \frac{P_Q}{P_{\text{total}}}=\frac8{13} \] \(\textbf{Physical interpretation:}\) The hotter section has greater resistance and therefore requires a greater voltage drop to maintain the common current. It also dissipates a greater fraction of the total electrical power. The voltage measurements reveal the temperature coefficient without requiring the absolute current or the wire's cross-sectional area. The result depends on treating the two sections as a series combination rather than as parallel conducting paths. \(\textbf{Final answer:}\) \[ \boxed{\alpha_0=6.0\times10^{-3}\,\text{K}^{-1},\quad\frac{P_Q}{P_{\text{total}}}=\frac8{13}} \]

Explanation: \(\textbf{Interpret the conducting geometry in experiment P:}\) Current travels between faces separated by \(1.0\,\text{cm}\). Therefore, the conducting length is \[ L_P=0.010\,\text{m} \] The perpendicular cross-sectional area is the product of the other two dimensions: \[ A_P=(0.020)(0.040) \] \[ A_P=8.0\times10^{-4}\,\text{m}^2 \] \(\textbf{Determine the measured resistance:}\) \[ R_P=\frac VI \] \[ R_P=\frac{12}{3.0} \] \[ R_P=4.0\,\Omega \] \(\textbf{Infer the material resistivity:}\) For uniform conduction, \[ R=\frac{\rho L}{A} \] Therefore, \[ \rho=\frac{RA}{L} \] Substituting, \[ \rho=\frac{(4.0)(8.0\times10^{-4})}{0.010} \] \[ \rho=0.32\,\Omega\,\text{m} \] \(\textbf{Identify the changed geometry:}\) In experiment Q, current travels across the \(4.0\,\text{cm}\) dimension. Thus, \[ L_Q=0.040\,\text{m} \] The perpendicular area is \[ A_Q=(0.010)(0.020) \] \[ A_Q=2.0\times10^{-4}\,\text{m}^2 \] \(\textbf{Calculate the second resistance:}\) \[ R_Q=\frac{\rho L_Q}{A_Q} \] \[ R_Q=\frac{(0.32)(0.040)}{2.0\times10^{-4}} \] \[ R_Q=64\,\Omega \] \(\textbf{Determine the new current:}\) \[ I_Q=\frac{12}{64} \] \[ I_Q=0.1875\,\text{A} \] \(\textbf{Compare the two orientations:}\) Moving the electrodes increases the conducting length by a factor of four. It also decreases the perpendicular area by a factor of four. The resistance therefore increases by a factor of sixteen. The material resistivity remains unchanged because neither the material nor its temperature has changed. The calculated current is consequently one-sixteenth of the original current. \(\textbf{Final answer:}\) \[ \boxed{\rho=0.32\,\Omega\,\text{m},\quad I_Q=0.1875\,\text{A}} \]

Explanation: \(\textbf{Use equal volume to compare areas:}\) For each uniform section, \[ \mathcal V=AL \] The volumes are equal, so \[ A_PL_P=A_QL_Q \] Given \(L_P=2L_Q\), \[ A_P=\frac{A_Q}{2} \] \(\textbf{Apply steady current continuity:}\) The sections form one unbranched conducting path. Therefore, their currents are equal: \[ I_P=I_Q=I \] Current density is \[ j=\frac IA \] Hence, \[ \frac{j_P}{j_Q}=\frac{A_Q}{A_P} \] \[ \frac{j_P}{j_Q}=2 \] \(\textbf{Compare electric-field magnitudes:}\) Microscopic Ohm's law gives \[ E=\rho j \] Therefore, \[ \frac{E_P}{E_Q} =\frac{\rho_P}{\rho_Q}\frac{j_P}{j_Q} \] Substituting, \[ \frac{E_P}{E_Q}=(2)(2) \] \[ \frac{E_P}{E_Q}=4 \] \(\textbf{Relate power density to current density:}\) The electrical heating rate per unit volume is \[ p=jE \] Using \(E=\rho j\), \[ p=\rho j^2 \] \(\textbf{Compare volumetric heating rates:}\) \[ \frac{p_P}{p_Q} =\frac{\rho_P}{\rho_Q}\left(\frac{j_P}{j_Q}\right)^2 \] \[ \frac{p_P}{p_Q}=2(2)^2 \] \[ \frac{p_P}{p_Q}=8 \] \(\textbf{Physical check:}\) Section P has greater resistivity and smaller cross-sectional area. Both properties increase the local electric field needed to maintain the common current. Because its current density is also greater, its heating per unit volume rises by a still larger factor. The heating-density ratio cannot be found from the resistivity ratio alone. It requires combining material behaviour with geometry and steady-current continuity. \(\textbf{Final answer:}\) \[ \boxed{\frac{E_P}{E_Q}=4,\quad\frac{p_P}{p_Q}=8} \]

Explanation: Statement I is correct because conductance is the reciprocal of resistance. The reciprocal of a positive quantity that changes linearly with temperature does not generally change linearly. Statement II is therefore incorrect. Equal temperature increases produce equal absolute resistance increases under the stipulated linear model, but they do not necessarily produce equal absolute conductance decreases. The conductance change depends on the resistance values at the beginning and end of each interval. Statement III is correct because conductivity and resistivity are reciprocal material properties. With unchanged geometry, wire resistance is proportional to resistivity. Consequently, decreasing conductivity increases resistivity and therefore resistance. Statement IV is correct because the ideal source maintains the same potential difference. The electrical power dissipated by the wire is \(P=V^2/R\). An increase in resistance thus reduces power. Over an equal fixed duration, electrical energy is power multiplied by time, so the energy dissipated also decreases. This trend would not be the same if the source instead maintained a fixed current. The question distinguishes intrinsic material properties, specimen conductance and the consequences of the external source constraint. Statements I, III and IV are correct.

Explanation: \(\textbf{Construct the resistance-temperature relation:}\) The resistance changes from \(5.0\,\Omega\) to \(7.0\,\Omega\) over \(100\,\text{K}\). Therefore, the slope is \[ \frac{\Delta R}{\Delta T}=\frac{7.0-5.0}{100} \] \[ \frac{\Delta R}{\Delta T}=0.020\,\Omega\,\text{K}^{-1} \] With \(T\) expressed numerically in degrees Celsius, \[ R_T=5.0+0.020T \] \(\textbf{Express the series current:}\) The thermometer has resistance \(R_T\), and the fixed resistor has resistance \(3.0\,\Omega\). Therefore, \[ I=\frac{12}{R_T+3.0} \] \(\textbf{Use the measured thermometer power:}\) The power dissipated in the thermometer is \[ P_T=I^2R_T \] Substituting the circuit current, \[ 9.0=\frac{12^2R_T}{(R_T+3.0)^2} \] Thus, \[ 9(R_T+3)^2=144R_T \] Dividing by nine, \[ (R_T+3)^2=16R_T \] Expanding, \[ R_T^2+6R_T+9=16R_T \] \[ R_T^2-10R_T+9=0 \] Factorising, \[ (R_T-1)(R_T-9)=0 \] \(\textbf{Apply the permissible temperature range:}\) The mathematical resistance solutions are \(1.0\,\Omega\) and \(9.0\,\Omega\). However, the calibrated temperature interval corresponds to resistances between \(5.0\,\Omega\) and \(7.0\,\Omega\). Neither mathematical solution lies in that interval. Therefore, the measured \(9.0\,\text{W}\) cannot occur under the stated temperature constraint. Indeed, throughout the allowed range the thermometer power is smaller than \(9.0\,\text{W}\). The measurements are incompatible with the permitted \(0^\circ ext{C}\) to \(100^\circ ext{C}\) interval; consequently, no temperature in that interval can satisfy both conditions.

Explanation: \(\textbf{Determine the mobility change:}\) In the elementary relaxation-time model, \[ \mu=\frac{e\tau}{m} \] The electron charge magnitude and effective mass are unchanged. Therefore, mobility is proportional to relaxation time: \[ \frac{\mu_2}{\mu_1}=\frac{\tau_2}{\tau_1} \] Given, \[ \tau_2=\frac{\tau_1}{3} \] Hence, \[ \frac{\mu_2}{\mu_1}=\frac13 \] \(\textbf{Determine the conductivity change:}\) For electron conduction, \[ \sigma=ne\mu \] Therefore, \[ \frac{\sigma_2}{\sigma_1} =\frac{n_2}{n_1}\frac{\mu_2}{\mu_1} \] Substituting, \[ \frac{\sigma_2}{\sigma_1}=2\left(\frac13\right) \] \[ \frac{\sigma_2}{\sigma_1}=\frac23 \] \(\textbf{Apply the equal-current-density requirement:}\) Microscopic Ohm's law gives \[ j=\sigma E \] The experiment requires \[ j_2=j_1 \] Therefore, \[ \sigma_2E_2=\sigma_1E_1 \] Rearranging, \[ \frac{E_2}{E_1}=\frac{\sigma_1}{\sigma_2} \] \[ \frac{E_2}{E_1}=\frac32 \] \(\textbf{Check the drift-speed change:}\) Equal current density also requires \[ n_2e|v_{d,2}|=n_1e|v_{d,1}| \] Because \(n_2=2n_1\), \[ \frac{|v_{d,2}|}{|v_{d,1}|}=\frac12 \] The mobility-field relation predicts \[ \frac{|v_{d,2}|}{|v_{d,1}|} =\frac{\mu_2}{\mu_1}\frac{E_2}{E_1} \] \[ \frac{|v_{d,2}|}{|v_{d,1}|} =\frac13\times\frac32=\frac12 \] The independent microscopic check agrees with the conductivity calculation. Doubling carrier density does not double conductivity when the mobility simultaneously decreases. The stronger field compensates for the reduced conductivity to preserve the specified current density. \(\textbf{Final answer:}\) \[ \boxed{\frac{\sigma_2}{\sigma_1}=\frac23,\quad\frac{E_2}{E_1}=\frac32} \]

Explanation: \(\textbf{Calculate signed charge in the first interval:}\) For steady current, \[ Q=It \] Thus, \[ Q_1=(+2.0)(3.0) \] \[ Q_1=+6.0\,\text{C} \] \(\textbf{Calculate signed charge in the second interval:}\) The current reverses direction. Therefore, \[ Q_2=(-3.0)(2.0) \] \[ Q_2=-6.0\,\text{C} \] \(\textbf{Net transported charge:}\) \[ Q_{\text{net}}=Q_1+Q_2 \] \[ Q_{\text{net}}=6.0-6.0 \] \[ Q_{\text{net}}=0\,\text{C} \] \(\textbf{Heating during the first interval:}\) For a resistor, \[ H=I^2Rt \] The first interval dissipates \[ H_1=(2.0)^2(2.0)(3.0) \] \[ H_1=24\,\text{J} \] \(\textbf{Heating during the reversed-current interval:}\) Electrical heating depends on the square of current, so reversing current does not reverse the sign of heating. \[ H_2=(-3.0)^2(2.0)(2.0) \] \[ H_2=36\,\text{J} \] \(\textbf{Total dissipated energy:}\) \[ H_{\text{total}}=24+36 \] \[ H_{\text{total}}=60\,\text{J} \] \(\textbf{Average power over the full interval:}\) \[ P_{\text{avg}}=\frac{H_{\text{total}}}{t_{\text{total}}} \] \[ P_{\text{avg}}=\frac{60}{5.0} \] \[ P_{\text{avg}}=12\,\text{W} \] The signed charge contributions cancel, but the electrical energies dissipated in the resistor add. Zero net transported charge therefore does not imply zero electrical heating. The distinction arises because charge transfer depends on the signed current, whereas Joule heating depends on its square. \(\textbf{Final answer:}\) \[ \boxed{Q_{\text{net}}=0\,\text{C},\quad P_{\text{avg}}=12\,\text{W}} \]

Explanation: \(\textbf{Connect charge transfer with heating:}\) For constant current, \[ Q=It \] Therefore, \[ I=\frac Qt \] Joule heating is \[ H=I^2Rt \] Substituting the charge relation, \[ H=\left(\frac Qt\right)^2Rt \] Thus, \[ H=\frac{Q^2R}{t} \] This relation is useful because the transported charge is specified independently of the current. \(\textbf{Determine the duration for P:}\) Rearranging, \[ t_P=\frac{Q^2R_P}{H_P} \] Substituting, \[ t_P=\frac{(12)^2(3.0)}{36} \] \[ t_P=12\,\text{s} \] \(\textbf{Determine the duration for Q:}\) \[ t_Q=\frac{Q^2R_Q}{H_Q} \] \[ t_Q=\frac{(12)^2(6.0)}{48} \] \[ t_Q=18\,\text{s} \] \(\textbf{Compare operating times:}\) \[ \frac{t_Q}{t_P}=\frac{18}{12} \] \[ \frac{t_Q}{t_P}=\frac32 \] \(\textbf{Determine average powers:}\) Each resistor operates at a constant current, so its average power is its total dissipated energy divided by operating time. For P, \[ P_P=\frac{36}{12} \] \[ P_P=3.0\,\text{W} \] For Q, \[ P_Q=\frac{48}{18} \] \[ P_Q=\frac83\,\text{W} \] \(\textbf{Compare average powers:}\) \[ \frac{P_Q}{P_P}=\frac{8/3}{3.0} \] \[ \frac{P_Q}{P_P}=\frac89 \] Although Q has twice the resistance of P, its current is smaller because the same charge is transported over a longer interval. Consequently, its average power is slightly smaller even though its total dissipated energy is greater. The fixed-charge condition changes how energy depends on operating duration. \(\textbf{Final answer:}\) \[ \boxed{\frac{t_Q}{t_P}=\frac32,\quad\frac{P_Q}{P_P}=\frac89} \]

Explanation: \(\textbf{First interval — Series arrangement:}\) The two resistors are connected in series. Their equivalent resistance is \[ R_s=6.0+6.0 \] \[ R_s=12\,\Omega \] \(\textbf{Source current during the first interval:}\) \[ I_s=\frac{12}{12} \] \[ I_s=1.0\,\text{A} \] \(\textbf{Energy dissipated during series operation:}\) The total series power is \[ P_s=VI_s \] \[ P_s=(12)(1.0) \] \[ P_s=12\,\text{W} \] For \(4.0\,\text{s}\), \[ H_s=P_st_s \] \[ H_s=(12)(4.0) \] \[ H_s=48\,\text{J} \] \(\textbf{Second interval — Parallel arrangement:}\) Two identical \(6.0\,\Omega\) resistors in parallel have equivalent resistance \[ R_p=\frac{6.0}{2} \] \[ R_p=3.0\,\Omega \] \(\textbf{Source current during parallel operation:}\) \[ I_p=\frac{12}{3.0} \] \[ I_p=4.0\,\text{A} \] \(\textbf{Energy dissipated during the second interval:}\) \[ P_p=VI_p \] \[ P_p=(12)(4.0) \] \[ P_p=48\,\text{W} \] Therefore, \[ H_p=(48)(2.0) \] \[ H_p=96\,\text{J} \] \(\textbf{Total dissipated electrical energy:}\) \[ H_{\text{total}}=H_s+H_p \] \[ H_{\text{total}}=48+96 \] \[ H_{\text{total}}=144\,\text{J} \] \(\textbf{Charge supplied by the source:}\) The source supplies different currents in the two intervals. Therefore, \[ Q_{\text{total}}=I_st_s+I_pt_p \] \[ Q_{\text{total}}=(1.0)(4.0)+(4.0)(2.0) \] \[ Q_{\text{total}}=12\,\text{C} \] The source voltage remains constant, but the electrical power changes when the resistor arrangement changes. Both charge and energy must be accumulated over the actual operating intervals. Using the final current for the entire experiment would overestimate both quantities. \(\textbf{Final answer:}\) \[ \boxed{H_{\text{total}}=144\,\text{J},\quad Q_{\text{total}}=12\,\text{C}} \]