Current Electricity MCQs with Answers – Part 5 (NEET Physics)

Explanation: \(\textbf{Identify the microscopic conduction relation:}\) For electron conduction, \[ j=ne\mu E \] The mobile-electron number density is unchanged between the operating conditions. Therefore, changes in the ratio \(j/E\) reveal changes in electron mobility. \(\textbf{Determine the first conductivity:}\) Microscopic Ohm's law gives \[ \sigma_1=\frac{j_1}{E_1} \] Substituting, \[ \sigma_1=\frac{4.0\times10^6}{1.0} \] \[ \sigma_1=4.0\times10^6\,\text{S m}^{-1} \] \(\textbf{Determine the second conductivity:}\) \[ \sigma_2=\frac{j_2}{E_2} \] \[ \sigma_2=\frac{8.0\times10^6}{3.0} \] \[ \sigma_2=\frac83\times10^6\,\text{S m}^{-1} \] \(\textbf{Infer the mobility ratio:}\) Because \[ \sigma=ne\mu \] and n is unchanged, \[ \frac{\mu_2}{\mu_1}=\frac{\sigma_2}{\sigma_1} \] Therefore, \[ \frac{\mu_2}{\mu_1} =\frac{(8/3)\times10^6}{4.0\times10^6} \] \[ \frac{\mu_2}{\mu_1}=\frac23 \] \(\textbf{Compare the drift speeds:}\) Current density is \[ j=ne|v_d| \] Thus, for constant n, \[ \frac{|v_{d,2}|}{|v_{d,1}|}=\frac{j_2}{j_1} \] \[ \frac{|v_{d,2}|}{|v_{d,1}|}=2 \] The drift speed doubles even though the mobility decreases. This is possible because the applied electric field becomes three times stronger. \(\textbf{Calculate the first heating-rate density:}\) Electrical power dissipated per unit volume is \[ p=jE \] Therefore, \[ p_1=(4.0\times10^6)(1.0) \] \[ p_1=4.0\times10^6\,\text{W m}^{-3} \] \(\textbf{Calculate the second heating-rate density:}\) \[ p_2=(8.0\times10^6)(3.0) \] \[ p_2=24\times10^6\,\text{W m}^{-3} \] Hence, \[ \frac{p_2}{p_1}=\frac{24}{4.0} \] \[ \frac{p_2}{p_1}=6 \] The current density increases by a factor of two, while the electric field increases by a factor of three. Their product therefore increases by a factor of six. Mobility, drift speed and heating rate respond differently because they depend on different combinations of carrier density and electric field. \(\textbf{Final answer:}\) \[ \boxed{\frac{\mu_2}{\mu_1}=\frac23,\quad\frac{p_2}{p_1}=6} \]

Explanation: \(\textbf{Convert measured resistance into conductivity:}\) For a uniform specimen, \[ R=\frac{L}{\sigma A} \] Its length and cross-sectional area remain unchanged. Therefore, \[ R\propto\frac1{\sigma} \] The resistance doubles: \[ \frac{R_2}{R_1}=2 \] Hence, \[ \frac{\sigma_2}{\sigma_1}=\frac12 \] \(\textbf{Use the independently measured mobility change:}\) The mobility ratio is \[ \frac{\mu_2}{\mu_1}=\frac14 \] For electron conduction, \[ \sigma=ne\mu \] Taking the ratio, \[ \frac{\sigma_2}{\sigma_1} =\frac{n_2}{n_1}\frac{\mu_2}{\mu_1} \] Substituting, \[ \frac12=\frac{n_2}{n_1}\left(\frac14\right) \] Therefore, \[ \frac{n_2}{n_1}=2 \] \(\textbf{Apply the constant-current source constraint:}\) The same current flows through the same cross-sectional area at both temperatures. For electron conduction, \[ I=neA|v_d| \] Therefore, \[ n_1|v_{d,1}|=n_2|v_{d,2}| \] Rearranging, \[ \frac{|v_{d,2}|}{|v_{d,1}|}=\frac{n_1}{n_2} \] Hence, \[ \frac{|v_{d,2}|}{|v_{d,1}|}=\frac12 \] \(\textbf{Check the required electric-field change:}\) With constant current and area, the current density is fixed. Microscopic Ohm's law gives \[ j=\sigma E \] Thus, \[ \frac{E_2}{E_1}=\frac{\sigma_1}{\sigma_2} \] \[ \frac{E_2}{E_1}=2 \] The mobility-field relation gives \[ \frac{|v_{d,2}|}{|v_{d,1}|} =\frac{\mu_2}{\mu_1}\frac{E_2}{E_1} \] \[ \frac{|v_{d,2}|}{|v_{d,1}|}=\frac14(2) \] \[ \frac{|v_{d,2}|}{|v_{d,1}|}=\frac12 \] The two microscopic approaches agree. The resistance measurement alone cannot determine whether the conductivity change arises from carrier density, mobility or both. The independent mobility measurement removes that ambiguity and allows the carrier density to be inferred. \(\textbf{Final answer:}\) \[ \boxed{\frac{n_2}{n_1}=2,\quad\frac{|v_{d,2}|}{|v_{d,1}|}=\frac12} \]

Explanation: \(\textbf{Determine the heated-wire resistance:}\) Microscopic conductivity is \[ \sigma=ne\mu \] The electron density is unchanged, while mobility is halved. Therefore, \[ \sigma_{\text{hot}}=\frac{\sigma_0}{2} \] Since the wire dimensions are unchanged, \[ R_{\text{hot}}=2R_0 \] \(\textbf{Experiment V — Fixed applied voltage:}\) The voltage is held at \(V_0\). The heated-wire current is \[ I_V=\frac{V_0}{2R_0} \] Using the original current \(I_0=V_0/R_0\), \[ I_V=\frac{I_0}{2} \] \(\textbf{Drift speed during fixed-voltage operation:}\) For unchanged n and A, \[ I=neA|v_d| \] Therefore, \[ |v_{d,V}|=\frac{|v_{d,0}|}{2} \] The electron drift traversal time consequently doubles: \[ t_V=\frac{L}{|v_{d,V}|} \] \[ t_V=2t_0 \] \(\textbf{Electrical power in experiment V:}\) \[ P_V=V_0I_V \] \[ P_V=V_0\left(\frac{I_0}{2}\right) \] \[ P_V=\frac{V_0I_0}{2} \] \(\textbf{Experiment I — Fixed current:}\) The ideal current source maintains \(I_0\). Therefore, \[ I_I=I_0 \] With unchanged carrier density and area, \[ |v_{d,I}|=|v_{d,0}| \] Thus, \[ t_I=t_0 \] \(\textbf{Compare drift traversal times:}\) \[ \frac{t_V}{t_I}=\frac{2t_0}{t_0} \] \[ \frac{t_V}{t_I}=2 \] \(\textbf{Determine the voltage required in experiment I:}\) The heated wire has resistance \(2R_0\). Therefore, \[ V_I=I_0(2R_0) \] \[ V_I=2V_0 \] \(\textbf{Electrical power in experiment I:}\) \[ P_I=V_II_I \] \[ P_I=(2V_0)I_0 \] \[ P_I=2V_0I_0 \] \(\textbf{Compare the heated-wire powers:}\) \[ \frac{P_I}{P_V} =\frac{2V_0I_0}{V_0I_0/2} \] \[ \frac{P_I}{P_V}=4 \] \(\textbf{Physical interpretation:}\) The two experiments use the same heated material with the same reduced mobility. However, their source constraints differ. Under fixed voltage, the current and drift speed decrease, so electrons require more time to traverse the wire. Under fixed current, the source increases its voltage enough to preserve drift speed, but the electrical heating rate rises. The ratio of powers is therefore four even though the wire's resistance is identical in the two heated experiments. \(\textbf{Final answer:}\) \[ \boxed{\frac{t_V}{t_I}=2,\quad\frac{P_I}{P_V}=4} \]

Explanation: \(\textbf{Establish the circuit operating condition:}\) The device and the fixed resistor are connected in series, so they carry the same current. The source voltage must equal the sum of their potential differences. Thus, \[ 10=V_D+2I_D \] This relation is the circuit's load condition and must be satisfied together with the measured device characteristic. \(\textbf{Test the first measured operating point:}\) At \(V_D=2.0\,\text{V}\), the device current is \(1.0\,\text{A}\). The required source voltage would be \[ V_s=2.0+(1.0)(2.0) \] \[ V_s=4.0\,\text{V} \] This is smaller than the available \(10\,\text{V}\). \(\textbf{Test the second measured operating point:}\) At \(V_D=4.0\,\text{V}\), the current is \(3.0\,\text{A}\). Therefore, \[ V_s=4.0+(3.0)(2.0) \] \[ V_s=10\,\text{V} \] This operating point satisfies the source-voltage condition exactly. \(\textbf{Calculate device power:}\) \[ P_D=V_DI_D \] \[ P_D=(4.0)(3.0) \] \[ P_D=12\,\text{W} \] \(\textbf{Calculate operating-point resistance:}\) \[ R_{\text{op}}=\frac{V_D}{I_D} \] \[ R_{\text{op}}=\frac{4.0}{3.0} \] \[ R_{\text{op}}=\frac43\,\Omega \] \(\textbf{Check the power balance:}\) The series resistor dissipates \[ P_R=I_D^2R \] \[ P_R=(3.0)^2(2.0)=18\,\text{W} \] The total resistive dissipation is \[ P_{\text{total}}=12+18=30\,\text{W} \] The ideal source supplies \[ P_s=(10)(3.0)=30\,\text{W} \] The two powers agree. The operating-point resistance is the ratio of the device voltage to its current at the selected point; it is not necessarily the slope of its nonlinear characteristic. The source voltage cannot be applied entirely across D because the series resistor also requires a voltage drop. \(\textbf{Final answer:}\) \[ \boxed{P_D=12\,\text{W},\quad R_{\text{op}}=\frac43\,\Omega} \]

Explanation: Statement I is incorrect because agreement between two measured resistance ratios does not establish the behaviour at every intermediate or additional operating point. A nonlinear characteristic can pass through two points having the same ratio \(V/I\) without being a straight line between them. Establishing ohmic behaviour requires proportionality between voltage and current over the relevant range under unchanged physical conditions. Statement II is correct because \(V/I\) can be calculated whenever the current is nonzero. For a non-ohmic device, this ratio describes its resistance at that particular operating point. It need not remain constant when the applied voltage changes. Statement III is correct because Ohm's law states that current is proportional to potential difference when the relevant physical conditions remain constant. A graph of V against I must consequently be a straight line through the origin, with slope equal to the constant resistance. Statement IV is incorrect because passing through the origin does not require every other point to lie on the same straight line. A nonlinear curve may pass through the origin while its ratio \(V/I\) varies from one nonzero point to another. The distinction is between a characteristic passing through one specified point and a characteristic having one constant slope throughout. The presence of a measurable voltage and current does not by itself establish Ohm's law. Only Statements II and III are correct.

Explanation: \(\textbf{Experiment S — Apply the common-current condition:}\) Devices connected in series must carry the same current. Their potential differences must add to the \(10\,\text{V}\) source voltage: \[ V_P+V_Q=10\,\text{V} \] At a current of \(2.0\,\text{A}\), the table gives \[ V_P=4.0\,\text{V} \] \[ V_Q=6.0\,\text{V} \] Therefore, \[ V_P+V_Q=4.0+6.0 \] \[ V_P+V_Q=10\,\text{V} \] The series operating current is \[ I_{\text{series}}=2.0\,\text{A} \] \(\textbf{Power dissipated in Q during series operation:}\) \[ P_{Q,S}=V_QI_{\text{series}} \] \[ P_{Q,S}=(6.0)(2.0) \] \[ P_{Q,S}=12\,\text{W} \] \(\textbf{Experiment P — Apply the common-voltage condition:}\) Devices connected in parallel experience the same terminal voltage. The applied voltage is \(6.0\,\text{V}\). From the table, P carries \[ I_P=3.0\,\text{A} \] at \(6.0\,\text{V}\). At the same voltage, Q carries \[ I_Q=2.0\,\text{A} \] \(\textbf{Determine the parallel source current:}\) \[ I_{\text{parallel}}=I_P+I_Q \] \[ I_{\text{parallel}}=3.0+2.0 \] \[ I_{\text{parallel}}=5.0\,\text{A} \] \(\textbf{Compare source currents:}\) \[ \frac{I_{\text{parallel}}}{I_{\text{series}}} =\frac{5.0}{2.0} \] \[ \frac{I_{\text{parallel}}}{I_{\text{series}}}=\frac52 \] The series experiment requires matching the devices at a common current and adding their voltages. The parallel experiment instead requires matching their voltages and adding their currents. These are different operations even when the same device characteristics are used. Assuming that both devices have one fixed resistance would ignore the non-ohmic behaviour of Q at higher currents. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{\text{parallel}}}{I_{\text{series}}}=\frac52,\quad P_{Q,S}=12\,\text{W}} \]

Explanation: \(\textbf{Distinguish the two ammeter positions:}\) In the first arrangement, the ammeter is upstream of the junction supplying the specimen and voltmeter. It therefore measures their combined current. In the second arrangement, the ammeter measures only the specimen current. The voltmeter remains connected across the same \(12\,\text{V}\), so its own current is unchanged. \(\textbf{Determine the voltmeter current:}\) Current conservation at the junction gives \[ I_{\text{total}}=I_{\text{specimen}}+I_V \] Substituting, \[ 3.0=2.0+I_V \] Therefore, \[ I_V=1.0\,\text{A} \] \(\textbf{Calculate the voltmeter resistance:}\) \[ R_V=\frac{V}{I_V} \] \[ R_V=\frac{12}{1.0} \] \[ R_V=12\,\Omega \] \(\textbf{Determine the specimen resistance:}\) The specimen current is \(2.0\,\text{A}\). Therefore, \[ R_s=\frac{12}{2.0} \] \[ R_s=6.0\,\Omega \] \(\textbf{Power consumed by the specimen:}\) \[ P_s=VI_s \] \[ P_s=(12)(2.0) \] \[ P_s=24\,\text{W} \] \(\textbf{Total power in the first arrangement:}\) Both parallel branches receive \(12\,\text{V}\). Therefore, \[ P_{\text{total}}=VI_{\text{total}} \] \[ P_{\text{total}}=(12)(3.0) \] \[ P_{\text{total}}=36\,\text{W} \] \(\textbf{Required power fraction:}\) \[ \frac{P_s}{P_{\text{total}}} =\frac{24}{36} \] \[ \frac{P_s}{P_{\text{total}}}=\frac23 \] The specimen itself is ohmic, but the finite-resistance voltmeter draws additional current. An ammeter placed before the junction includes that current and therefore cannot be treated as measuring the specimen current alone. Comparing the two readings separates the instrument-loading current from the specimen current. The two power contributions must likewise be distinguished. \(\textbf{Final answer:}\) \[ \boxed{R_V=12\,\Omega,\quad\frac{P_s}{P_{\text{total}}}=\frac23} \]

Explanation: \(\textbf{Identify the lamp's operating point:}\) The lamp and \(2.0\,\Omega\) resistor carry a common current. Their voltage drops must add to the source voltage: \[ V_L+2I=8.0 \] At the measured lamp point \(V_L=6.0\,\text{V}\) and \(I=1.0\,\text{A}\), \[ 6.0+2(1.0)=8.0 \] The operating point satisfies the circuit condition. Therefore, \[ I_L=1.0\,\text{A} \] \(\textbf{Power dissipated in the filament lamp:}\) \[ P_L=V_LI_L \] \[ P_L=(6.0)(1.0) \] \[ P_L=6.0\,\text{W} \] \(\textbf{Determine the specified replacement resistance:}\) At the lamp's \(2.0\,\text{V}\) operating point, its resistance ratio is \[ R_{\text{op},2}=\frac{V_L}{I_L} \] \[ R_{\text{op},2}=\frac{2.0}{0.50} \] \[ R_{\text{op},2}=4.0\,\Omega \] The replacement resistor is ohmic and therefore retains this resistance. \(\textbf{Current with the replacement resistor:}\) The complete series resistance becomes \[ R_{\text{total}}=4.0+2.0 \] \[ R_{\text{total}}=6.0\,\Omega \] Thus, \[ I_R=\frac{8.0}{6.0} \] \[ I_R=\frac43\,\text{A} \] \(\textbf{Power dissipated in the replacement resistor:}\) \[ P_R=I_R^2(4.0) \] \[ P_R=\left(\frac43\right)^2(4.0) \] \[ P_R=\frac{64}{9}\,\text{W} \] \(\textbf{Compare the device powers:}\) \[ \frac{P_L}{P_R} =\frac{6.0}{64/9} \] \[ \frac{P_L}{P_R}=\frac{54}{64} \] \[ \frac{P_L}{P_R}=\frac{27}{32} \] The filament lamp does not retain its \(2.0\,\text{V}\) resistance when operating at \(6.0\,\text{V}\). Its steady-state resistance increases as its filament becomes hotter. The replacement resistor instead has a fixed \(4.0\,\Omega\) resistance. Both devices are connected to the same source arrangement, but their different operating characteristics lead to different currents and powers. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_L}{P_R}=\frac{27}{32},\quad I_L=1.0\,\text{A}} \]

Explanation: The Assertion is true under the stated experimental conditions. Ohm's law requires the relevant physical conditions of a conductor to remain unchanged. Temperature is one of those conditions because the resistance of an ordinary metallic filament increases as its temperature rises. Under continuous DC operation, electrical heating can raise the filament temperature as the applied voltage increases. Consequently, the filament resistance at a higher steady operating current may differ from its resistance at a lower current. The resulting steady-state voltage and current readings need not form a straight line through the origin. However, sufficiently brief measurements can reduce the temperature change occurring during each observation. If the filament begins at the same temperature and remains approximately at that temperature during each brief measurement, its resistance can remain approximately constant. Voltage and current may then be approximately proportional over the tested range. The Reason correctly explains the difference between the two experimental characteristics. A curved steady-state characteristic does not establish that Ohm's law is violated when the conductor's temperature is held fixed. It may instead show that the physical state of the conductor changes between measurements. Both A and R are true, and R correctly explains A.

Explanation: \(\textbf{Separate true current from displayed current:}\) The resistor is ohmic at constant temperature. Therefore, \[ I_{\text{true}}=\frac VR \] The ammeter adds a constant zero error b: \[ I_{\text{display}}=\frac VR+b \] This produces a straight-line graph of displayed current against true voltage, but the line need not pass through the origin. \(\textbf{Use the first observation:}\) \[ 1.20=\frac{6.0}{R}+b \] \(\textbf{Use the second observation:}\) \[ 3.20=\frac{18}{R}+b \] \(\textbf{Eliminate the zero error:}\) Subtracting the first equation from the second, \[ 3.20-1.20=\frac{18-6.0}{R} \] \[ 2.0=\frac{12}{R} \] Therefore, \[ R=6.0\,\Omega \] \(\textbf{Recover the zero error:}\) The true current at \(6.0\,\text{V}\) is \[ I_{\text{true},1}=\frac{6.0}{6.0} \] \[ I_{\text{true},1}=1.0\,\text{A} \] Hence, \[ b=1.20-1.0 \] \[ b=0.20\,\text{A} \] \(\textbf{Check the second reading:}\) At \(18\,\text{V}\), the true current is \[ I_{\text{true},2}=\frac{18}{6.0} \] \[ I_{\text{true},2}=3.0\,\text{A} \] Adding the zero error gives \[ I_{\text{display},2}=3.0+0.20 \] \[ I_{\text{display},2}=3.20\,\text{A} \] Both observations are reproduced. The displayed ratio \(V/I_{\text{display}}\) is not the actual resistance because the ammeter readings contain an additive error. Taking the difference between measurements removes that error. The corrected voltage-current characteristic passes through the origin, consistent with an ohmic resistor. \(\textbf{Final answer:}\) \[ \boxed{R=6.0\,\Omega,\quad b=0.20\,\text{A}} \]

Explanation: \(\textbf{Identify the operating point:}\) At an applied voltage of \(4.0\,\text{V}\), the measured current is \(4.0\,\text{A}\). Operating-point resistance is the ratio of the voltage to the current at that specific point: \[ R_{\text{op}}=\frac VI \] Substituting, \[ R_{\text{op}}=\frac{4.0}{4.0} \] \[ R_{\text{op}}=1.0\,\Omega \] \(\textbf{Determine the changes across the specified interval:}\) The voltage rises from \(4.0\,\text{V}\) to \(6.0\,\text{V}\). Therefore, \[ \Delta V=6.0-4.0 \] \[ \Delta V=2.0\,\text{V} \] The current rises from \(4.0\,\text{A}\) to \(9.0\,\text{A}\): \[ \Delta I=9.0-4.0 \] \[ \Delta I=5.0\,\text{A} \] \(\textbf{Calculate the average incremental resistance:}\) \[ R_{\text{inc}}=\frac{\Delta V}{\Delta I} \] \[ R_{\text{inc}}=\frac{2.0}{5.0} \] \[ R_{\text{inc}}=\frac25\,\Omega \] \(\textbf{Interpret the two resistance values:}\) The operating-point resistance uses the coordinates of one point relative to the origin. The average incremental resistance uses the changes between two specified points. For a straight V-I characteristic through the origin, these quantities would be equal. For the present nonlinear characteristic, they are different. The larger current response between \(4.0\,\text{V}\) and \(6.0\,\text{V}\) corresponds to a smaller incremental resistance. The device cannot be replaced by a single constant resistance that reproduces every measurement. \(\textbf{Final answer:}\) \[ \boxed{R_{\text{op}}=1.0\,\Omega,\quad R_{\text{inc}}=\frac25\,\Omega} \]

Explanation: \(\textbf{Apply the common-voltage condition:}\) The non-ohmic device and ohmic resistor are connected in parallel. Their potential differences are equal, even though their currents may differ. The ideal current source maintains the sum of their currents: \[ I_{\text{source}}=I_D+I_R \] The supplied current is \(9.0\,\text{A}\). \(\textbf{Express current through the ohmic resistor:}\) For the \(2.0\,\Omega\) resistor, \[ I_R=\frac V{2.0} \] The device current must be taken from its measured characteristic at the same voltage V. \(\textbf{Test the device point at }4.0\,\textbf{V:}\) At \(4.0\,\text{V}\), D carries \(3.0\,\text{A}\). The resistor would carry \[ I_R=\frac{4.0}{2.0}=2.0\,\text{A} \] The combined current would be \[ I_{\text{total}}=3.0+2.0 \] \[ I_{\text{total}}=5.0\,\text{A} \] This is below the required \(9.0\,\text{A}\). \(\textbf{Test the device point at }6.0\,\textbf{V:}\) At \(6.0\,\text{V}\), D carries \(6.0\,\text{A}\). The resistor carries \[ I_R=\frac{6.0}{2.0} \] \[ I_R=3.0\,\text{A} \] The source current is therefore \[ I_{\text{total}}=6.0+3.0 \] \[ I_{\text{total}}=9.0\,\text{A} \] This is the specified operating condition. Hence, \[ V=6.0\,\text{V} \] \(\textbf{Determine D's current fraction:}\) \[ \frac{I_D}{I_{\text{source}}}=\frac{6.0}{9.0} \] \[ \frac{I_D}{I_{\text{source}}}=\frac23 \] \(\textbf{Check branch powers:}\) The device dissipates \[ P_D=VI_D=(6.0)(6.0)=36\,\text{W} \] The ohmic resistor dissipates \[ P_R=(6.0)(3.0)=18\,\text{W} \] Their total \(54\,\text{W}\) agrees with the source power \((6.0)(9.0)\). An ideal current source sets total current rather than total voltage. The operating voltage emerges from the sum of the two parallel branch characteristics. The current cannot simply be divided according to fixed resistances because D is non-ohmic. \(\textbf{Final answer:}\) \[ \boxed{V=6.0\,\text{V},\quad\frac{I_D}{I_{\text{source}}}=\frac23} \]

Explanation: \(\textbf{Observation P — slope of voltage against current:}\) Observation P concerns a graph of voltage against current. For an ohmic conductor, the slope of that graph is resistance because \(V=IR\). \[ R_P=\frac{\Delta V}{\Delta I}=3.0\,\Omega \] A straight line through the origin with slope \(3.0\,\text{V A}^{-1}\) therefore represents a \(3.0\,\Omega\) resistor. Thus, P matches 2. \(\textbf{Observation Q — one point on a nonlinear characteristic:}\) Observation Q concerns one particular point on a nonlinear characteristic. At that point, the ratio \(V/I\) is \[ R_{\text{operating point}}=\frac{8.0}{2.0}=4.0\,\Omega \] However, knowing this ratio at one point does not establish constant resistance at other currents. Thus, Q matches 3. \(\textbf{Observation R — the axes are reversed:}\) The slope of an I-against-V graph is current divided by voltage, which equals conductance for an ohmic conductor. \[ R_R=\frac1{0.50}=2.0\,\Omega \] Thus, R matches 1. \(\textbf{Observation S — compare two operating points:}\) For observation S, the current increases by a greater factor than the applied voltage. Since operating-point resistance is \(V/I\), that resistance decreases between the two operating points. Thus, S matches 4. The matching requires keeping track of graph-axis orientation and distinguishing operating-point resistance from constant ohmic resistance. The complete matching is P–2, Q–3, R–1 and S–4.

Explanation: \(\textbf{Determine the series operating current:}\) In experiment S, the same current passes through P and Q. The voltage across the ohmic resistor is \[ V_P=3I \] The source condition is \[ 12=3I+V_Q \] At \(I=2.0\,\text{A}\), the measured voltage across Q is \(6.0\,\text{V}\). Therefore, \[ 3(2.0)+6.0=12\,\text{V} \] The series operating current is \[ I_{\text{series}}=2.0\,\text{A} \] \(\textbf{Power in Q during series operation:}\) \[ P_{Q,S}=V_QI \] \[ P_{Q,S}=(6.0)(2.0) \] \[ P_{Q,S}=12\,\text{W} \] \(\textbf{Determine parallel branch currents:}\) In experiment P, both devices receive \(6.0\,\text{V}\). The current through the ohmic resistor is \[ I_P=\frac{6.0}{3.0} \] \[ I_P=2.0\,\text{A} \] At \(6.0\,\text{V}\), the non-ohmic device also carries \[ I_Q=2.0\,\text{A} \] \(\textbf{Parallel source current:}\) \[ I_{\text{parallel}}=I_P+I_Q \] \[ I_{\text{parallel}}=2.0+2.0 \] \[ I_{\text{parallel}}=4.0\,\text{A} \] Thus, \[ \frac{I_{\text{parallel}}}{I_{\text{series}}} =\frac{4.0}{2.0} \] \[ \frac{I_{\text{parallel}}}{I_{\text{series}}}=2 \] \(\textbf{Power in Q during parallel operation:}\) \[ P_{Q,P}=V_QI_Q \] \[ P_{Q,P}=(6.0)(2.0) \] \[ P_{Q,P}=12\,\text{W} \] Therefore, \[ \frac{P_{Q,P}}{P_{Q,S}}=\frac{12}{12} \] \[ \frac{P_{Q,P}}{P_{Q,S}}=1 \] Although the source connections and source voltages differ, Q operates at the same voltage and current in both experiments. Its individual power therefore remains unchanged. The parallel source supplies twice the series source current because it also supplies the separate ohmic branch. This illustrates why source-current comparisons alone do not determine the operating power of an individual nonlinear device. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_{\text{parallel}}}{I_{\text{series}}}=2,\quad\frac{P_{Q,P}}{P_{Q,S}}=1} \]

Explanation: \(\textbf{Identify the limiting device condition:}\) The current through D must not exceed \(2.0\,\text{A}\). Because D is in series with R, the series current is also the device current. The smallest safe resistance occurs when the current reaches, but does not exceed, this limit. \(\textbf{Read the voltage required by D at its current limit:}\) At \[ I_{\max}=2.0\,\text{A}, \] the table gives \[ V_D=4.0\,\text{V} \] \(\textbf{Determine the voltage across the adjustable resistor:}\) The source maintains \(12\,\text{V}\). The two series voltage drops must add to this value. Thus, \[ V_R=V_s-V_D \] \[ V_R=12-4.0 \] \[ V_R=8.0\,\text{V} \] \(\textbf{Calculate the limiting series resistance:}\) At the current limit, \[ R_{\min}=\frac{V_R}{I_{\max}} \] Substituting, \[ R_{\min}=\frac{8.0}{2.0} \] \[ R_{\min}=4.0\,\Omega \] \(\textbf{Verify that a smaller resistance would violate the limit:}\) The source condition is \[ 12=V_D(I)+IR \] The device voltage increases with its current. If R is made smaller while the current is held at \(2.0\,\text{A}\), the total required voltage becomes less than \(12\,\text{V}\). The operating current must then increase to satisfy the source condition. Therefore, any smaller R would exceed the prescribed limit. \(\textbf{Device power at the limiting setting:}\) \[ P_D=V_DI_{\max} \] \[ P_D=(4.0)(2.0) \] \[ P_D=8.0\,\text{W} \] \(\textbf{Check the complete circuit power:}\) The series resistor dissipates \[ P_R=I_{\max}^2R_{\min} \] \[ P_R=(2.0)^2(4.0) \] \[ P_R=16\,\text{W} \] The source supplies \[ P_s=(12)(2.0)=24\,\text{W} \] The sum \(8.0+16=24\,\text{W}\) confirms the energy balance. The limiting resistance must be calculated using the nonlinear device's actual voltage at the maximum permitted current. Treating the device as a fixed resistor based on an unrelated operating point could produce an unsafe design. \(\textbf{Final answer:}\) \[ \boxed{R_{\min}=4.0\,\Omega,\quad P_D=8.0\,\text{W}} \]

Explanation: Statement I is correct because the current magnitudes are different for equal-magnitude voltages of opposite polarity. At \(+2.0\,\text{V}\), the current magnitude is \(0.50\,\text{A}\), while at \(-2.0\,\text{V}\), it is only \(0.010\,\text{A}\). The device therefore has an asymmetric voltage-current characteristic. Statement II is incorrect because the signs of voltage and current must be interpreted together under the passive sign convention. Power absorbed by the device is \(P=VI\). In the reverse-polarity measurement, both \(V\) and \(I\) are negative, so their product is positive. The device absorbs rather than supplies electrical power. \(\textbf{Compare the operating-point resistances:}\) Statement III is correct because operating-point resistance magnitude is \(|V|/|I|\). \[ |R_+|=\frac{2.0}{0.50}=4.0\,\Omega \] \[ |R_-|=\frac{2.0}{0.010}=200\,\Omega \] The ratio is \[ \frac{|R_-|}{|R_+|}=\frac{200}{4.0}=50 \] \(\textbf{Compare electrical powers:}\) Statement IV is correct because voltage and current have the same sign in each measurement. The forward-polarity power is \[ P_+=(+2.0)(+0.50)=1.0\,\text{W} \] The reverse-polarity power is \[ P_-=(-2.0)(-0.010)=+0.020\,\text{W} \] Both powers represent electrical energy absorbed by the passive device. The opposite current signs describe opposite charge-flow directions, not automatically opposite directions of energy transfer. Thus, Statements I, III and IV are correct.

Explanation: \(\textbf{Identify the requested operating point:}\) The graph gives \[ I=2.0\,\text{A} \] \[ V=7.0\,\text{V} \] The operating-point resistance is \[ R_{\text{op}}=\frac VI \] Therefore, \[ R_{\text{op}}=\frac{7.0}{2.0} \] \[ R_{\text{op}}=3.5\,\Omega \] \(\textbf{Determine the first interval's average incremental resistance:}\) From \(1.0\,\text{A}\) to \(2.0\,\text{A}\), the voltage increases from \(4.0\,\text{V}\) to \(7.0\,\text{V}\). Thus, \[ R_{\text{inc},1} =\frac{7.0-4.0}{2.0-1.0} \] \[ R_{\text{inc},1}=3.0\,\Omega \] \(\textbf{Determine the second interval's average incremental resistance:}\) From \(2.0\,\text{A}\) to \(3.0\,\text{A}\), the voltage increases from \(7.0\,\text{V}\) to \(9.0\,\text{V}\). Therefore, \[ R_{\text{inc},2} =\frac{9.0-7.0}{3.0-2.0} \] \[ R_{\text{inc},2}=2.0\,\Omega \] \(\textbf{Compare the increments:}\) \[ \frac{R_{\text{inc},1}}{R_{\text{inc},2}} =\frac{3.0}{2.0} \] \[ \frac{R_{\text{inc},1}}{R_{\text{inc},2}}=\frac32 \] \(\textbf{Interpret the changing graph slope:}\) A smaller V-against-I slope means that a given additional current requires a smaller additional voltage over the corresponding interval. The average incremental resistance decreases from \(3.0\,\Omega\) to \(2.0\,\Omega\). The operating-point resistance at \(2.0\,\text{A}\) is nevertheless \(3.5\,\Omega\). These values need not coincide because the graph is not a straight line through the origin. The operating-point ratio describes the whole voltage-current pair at one point, while each incremental ratio describes the change between two points. A nonlinear characteristic must therefore be interpreted according to the exact resistance definition used in the question. \(\textbf{Final answer:}\) \[ \boxed{R_{\text{op}}=3.5\,\Omega,\quad\frac{R_{\text{inc},1}}{R_{\text{inc},2}}=\frac32} \]

Explanation: \(\textbf{Determine the initial source current:}\) Initially, only P and Q are connected. Their resistances add: \[ R_i=3.0+6.0 \] \[ R_i=9.0\,\Omega \] Therefore, \[ I_i=\frac{15}{9.0} \] \[ I_i=\frac53\,\text{A} \] \(\textbf{Identify the modified current paths:}\) After D is connected, it is parallel to Q. Both therefore experience the same voltage \(V_p\). The source current flows through P and then divides between Q and D: \[ I_s=I_Q+I_D \] For the ohmic branch, \[ I_Q=\frac{V_p}{6.0} \] The current through D must be obtained from its nonlinear characteristic at the same voltage. \(\textbf{Write the source-voltage condition:}\) The source voltage is divided between P and the parallel section: \[ 15=3.0I_s+V_p \] Substituting the junction-current relation, \[ 15=3.0\left(\frac{V_p}{6.0}+I_D\right)+V_p \] \(\textbf{Test the device point at }6.0\,\textbf{V:}\) At \(V_p=6.0\,\text{V}\), D carries \(2.0\,\text{A}\). The current through Q is \[ I_Q=\frac{6.0}{6.0} \] \[ I_Q=1.0\,\text{A} \] The resulting source current is \[ I_f=1.0+2.0 \] \[ I_f=3.0\,\text{A} \] \(\textbf{Verify the voltage balance:}\) The voltage drop across P is \[ V_P=I_fR_P \] \[ V_P=(3.0)(3.0) \] \[ V_P=9.0\,\text{V} \] The source voltage is \[ V_P+V_p=9.0+6.0 \] \[ V_P+V_p=15\,\text{V} \] The measured device point therefore satisfies both current conservation and the loop voltage condition. \(\textbf{Compare source currents:}\) \[ \frac{I_f}{I_i}=\frac{3.0}{5/3} \] \[ \frac{I_f}{I_i}=\frac95 \] \(\textbf{Calculate power dissipated in D:}\) \[ P_D=V_pI_D \] \[ P_D=(6.0)(2.0) \] \[ P_D=12\,\text{W} \] \(\textbf{Check the complete power balance:}\) P dissipates \[ P_P=(3.0)^2(3.0)=27\,\text{W} \] Q dissipates \[ P_Q=\frac{6.0^2}{6.0}=6.0\,\text{W} \] The total load power is \[ 27+6.0+12=45\,\text{W} \] The ideal source supplies \[ P_s=(15)(3.0)=45\,\text{W} \] The powers agree. Adding D increases the source current while reducing the voltage across the original resistor Q. Because D is non-ohmic, its current must be read from the characteristic at the actual shared branch voltage. The operating point cannot be found by assuming that D has one constant resistance independent of applied voltage. \(\textbf{Final answer:}\) \[ \boxed{\frac{I_f}{I_i}=\frac95,\quad P_D=12\,\text{W}} \]

Explanation: \(\textbf{Identify the original current paths:}\) Before the wire is connected, current can travel from A to C along two separate paths. The path through B contains the \(2.0\,\Omega\) and \(6.0\,\Omega\) resistors in series. The path through D contains the two \(4.0\,\Omega\) resistors in series. \(\textbf{Reduce the original network:}\) \[ R_{ABC}=2.0+6.0=8.0\,\Omega \] \[ R_{ADC}=4.0+4.0=8.0\,\Omega \] These two paths are parallel. Therefore, \[ R_i=\frac{(8.0)(8.0)}{8.0+8.0} \] \[ R_i=4.0\,\Omega \] \(\textbf{Calculate the initial source current:}\) \[ I_i=\frac{28}{4.0} \] \[ I_i=7.0\,\text{A} \] \(\textbf{Identify the topology after joining B and D:}\) The ideal wire makes B and D one electrical node. The \(2.0\,\Omega\) resistor from A to B and the \(4.0\,\Omega\) resistor from A to D now connect between the same nodes. They form one parallel section. Similarly, the \(6.0\,\Omega\) resistor from B to C and the \(4.0\,\Omega\) resistor from D to C form another parallel section. The two parallel sections are connected in series. \(\textbf{Reduce the section adjacent to A:}\) \[ R_{A}=\frac{(2.0)(4.0)}{2.0+4.0} \] \[ R_A=\frac43\,\Omega \] \(\textbf{Reduce the section adjacent to C:}\) \[ R_C=\frac{(6.0)(4.0)}{6.0+4.0} \] \[ R_C=\frac{12}{5}\,\Omega \] \(\textbf{Determine the new equivalent resistance:}\) \[ R_f=\frac43+\frac{12}{5} \] \[ R_f=\frac{20+36}{15} \] \[ R_f=\frac{56}{15}\,\Omega \] \(\textbf{Determine the final source current:}\) \[ I_f=\frac{28}{56/15} \] \[ I_f=\frac{15}{2}\,\text{A} \] \(\textbf{Voltage across the B–C and D–C parallel section:}\) \[ V_C=I_fR_C \] \[ V_C=\frac{15}{2}\times\frac{12}{5} \] \[ V_C=18\,\text{V} \] \(\textbf{Current through the }6.0\,\Omega\textbf{ resistor:}\) \[ I_6=\frac{18}{6.0} \] \[ I_6=3.0\,\text{A} \] \(\textbf{Compare source currents:}\) \[ \frac{I_f}{I_i}=\frac{15/2}{7.0} \] \[ \frac{I_f}{I_i}=\frac{15}{14} \] The added wire creates new parallel combinations without changing any resistor value. The source current increases slightly, but the current through the \(6.0\,\Omega\) resistor decreases from its original value of \(3.5\,\text{A}\). Thus, an increase in total source current does not require every resistor current to increase. \(\textbf{Final answer:}\) \[ \boxed{I_6=3.0\,\text{A},\quad\frac{I_f}{I_i}=\frac{15}{14}} \]

Explanation: \(\textbf{Recover the unknown resistance from the original chain:}\) Initially, the three resistors form one unbranched path. Their equivalent resistance is their sum: \[ R_{AD}=2.0+X+4.0 \] Using the measured resistance, \[ 12=6.0+X \] Therefore, \[ X=6.0\,\Omega \] \(\textbf{Identify the new electrical nodes:}\) Connecting B directly to D makes those terminals equipotential. The unknown resistor X connects B to C. The \(4.0\,\Omega\) resistor connects C to D. Since B and D are now one node, X and the \(4.0\,\Omega\) resistor are parallel. The \(2.0\,\Omega\) resistor remains in series with this parallel section. \(\textbf{Reduce the parallel section:}\) \[ R_p=\frac{(6.0)(4.0)}{6.0+4.0} \] \[ R_p=\frac{12}{5}\,\Omega \] \(\textbf{Calculate the resistance across the final source:}\) \[ R_{\text{eq}}=2.0+\frac{12}{5} \] \[ R_{\text{eq}}=\frac{22}{5}\,\Omega \] \(\textbf{Determine the source current:}\) \[ I_s=\frac{22}{22/5} \] \[ I_s=5.0\,\text{A} \] \(\textbf{Calculate the voltage across X:}\) The parallel section has potential difference \[ V_p=I_sR_p \] \[ V_p=(5.0)\left(\frac{12}{5}\right) \] \[ V_p=12\,\text{V} \] Both parallel resistors receive this same voltage. \(\textbf{Determine current through X:}\) \[ I_X=\frac{V_p}{X} \] \[ I_X=\frac{12}{6.0} \] \[ I_X=2.0\,\text{A} \] \(\textbf{Calculate the requested power:}\) \[ P_X=I_X^2X \] \[ P_X=(2.0)^2(6.0) \] \[ P_X=24\,\text{W} \] \(\textbf{Check current conservation:}\) The \(4.0\,\Omega\) resistor carries \[ I_4=\frac{12}{4.0}=3.0\,\text{A} \] Its current and the current through X add to the source current: \[ 2.0+3.0=5.0\,\text{A} \] The B–D connection changes which resistors share the same pair of electrical nodes. Consequently, the original series resistance of \(12\,\Omega\) cannot be used as the final equivalent resistance. The topology must be reconstructed before the power in X is calculated. \(\textbf{Final answer:}\) \[ \boxed{X=6.0\,\Omega,\quad P_X=24\,\text{W}} \]