Physics 2 Quiz: Ohms Law And Power

What this quiz covers

This quiz focuses on Ohms Law And Power, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\dfrac{1}{7}$, because the internal resistance is $\frac{1}{7}$ of the total resistance in the circuit, and power dissipated is proportional to resistance when current is the same through all elements. (correct answer)
2. $\dfrac{1}{14}$, because the voltage drop across the internal resistance is $\frac{1}{14}$ of the EMF, and the power ratio equals the voltage ratio for series elements.
3. $\dfrac{1}{4}$, because the internal resistance $r = 2\,\Omega$ is compared to the external resistance of $8\,\Omega$, giving a ratio of $\frac{1}{4}$ of the externally delivered power.
4. $\dfrac{2}{7}$, because the current through the internal resistance is twice the current through either external resistor, doubling the effective power fraction beyond the resistance ratio.

Question 2

1. $550\,\text{W}$, because the total input power is $P_{\text{in}} = VI = 600\,\text{W}$, and the power lost to heat in the coil resistance is $P_R = I^2R = 50\,\text{W}$, leaving $550\,\text{W}$ for mechanical output. (correct answer)
2. $600\,\text{W}$, because all of the power delivered by the source, $P = VI = (120)(5) = 600\,\text{W}$, is converted to mechanical work since an ideal motor has no resistive losses.
3. $500\,\text{W}$, because the back-EMF of the motor is $\mathcal{E}_{\text{back}} = V - IR = 120 - 10 = 110\,\text{V}$, and the mechanical power is $P = \mathcal{E}_{\text{back}} \cdot I = 110 \times 5 = 550\,\text{W}$ minus the 50 W lost to heat gives 500 W.
4. $50\,\text{W}$, because the only useful work comes from the current flowing through the coil resistance, calculated as $P = I^2R = (5)^2(2) = 50\,\text{W}$, while the remaining power maintains the back-EMF.

Question 3

1. The device is ohmic with $R = 4\,\Omega$, and the power at $4\,\text{V}$ is $4\,\text{W}$, because the ratio $V/I$ is approximately constant across all three data points.
2. The device is non-ohmic because the ratio $V/I$ is not constant; $I \propto V^2$ fits the data, and the power at $4\,\text{V}$ is $8\,\text{W}$, calculated directly as $P = IV = (2)(4)$. (correct answer)
3. The device is non-ohmic because the ratio $V/I$ is not constant; $I \propto V^2$ fits the data, and the power at $4\,\text{V}$ is $4\,\text{W}$, because applying $P = I^2/R$ with an effective resistance of $2\,\Omega$ gives this result.
4. The device is non-ohmic with an effective resistance that decreases with voltage; the power at $4\,\text{V}$ is $16\,\text{W}$, because $P = V^2/R_{\text{eff}}$ and $R_{\text{eff}} = 1\,\Omega$ at that operating point.

Question 4

1. $P_C : P_A = 2 : 1$, because $R_C$ carries twice the current of $R_A$, and students who forget that power scales as the square of current — not linearly — arrive at twice the power rather than four times.
2. $P_C : P_A = 1 : 1$, because all three resistors are identical and the circuit is symmetric, so each must dissipate the same fraction of the total power from the battery.
3. $P_C : P_A = 4 : 1$, because $R_C$ carries the full circuit current $I$ while $R_A$ carries only $I/2$, and since $P \propto I^2R$ with equal resistances, the ratio is $(I)^2 R \,:\, (I/2)^2 R = 4 : 1$. (correct answer)
4. $P_C : P_A = 1 : 4$, because the voltage across the parallel combination ($\mathcal{E}/3$) is half the voltage across $R_C$ ($2\mathcal{E}/3$), and students who invert the power ratio — attributing higher power to the element with lower voltage — arrive at 1:4.

Question 5

1. The power decreases by a factor of 2, because doubling the length doubles the resistance while the area is unchanged, and $P = V^2/R$ therefore halves.
2. The power increases by a factor of 2, because the new resistance is $R' = \rho(2L)/[\pi(2r)^2] = \rho(2L)/(4\pi r^2) = R/2$, so with constant voltage $P = V^2/R$ gives $P' = 2V^2/R = 2P$. (correct answer)
3. The power decreases by a factor of 2, because the new resistance is $R' = R/2$, and applying $P = I^2 R$ with the original current $I$ gives $P' = I^2(R/2) = P/2$.
4. The power remains the same, because doubling both the length and the radius preserves the aspect ratio $L/r$, and students often (incorrectly) associate a preserved geometric ratio with a preserved resistance and therefore unchanged power.

Question 6

1. Approximately 10%, because power is proportional to voltage and a 10% decrease in voltage produces a 10% decrease in power.
2. Approximately 19%, because power is proportional to $V^2$, and a 10% decrease in voltage gives $(0.9)^2 = 0.81$, representing a 19% decrease in power. (correct answer)
3. Approximately 20%, because the current also decreases by 10%, and since $P = IV$, the combined 10% decrease in both $I$ and $V$ produces a 20% decrease — but this must be corrected for the nonlinearity, yielding approximately 19%.
4. Approximately 21%, because the resistance of a heating element slightly decreases as it cools during reduced power operation, which amplifies the power reduction beyond the simple $(0.9)^2$ prediction.

Question 7

1. $16\,\text{W}$, because the terminal voltage across the parallel combination is $V_T = \mathcal{E} - I \cdot r = 20 - (2)(2) = 16\,\text{V}$, and power is $P = V_T / R_{\text{ext}} = 16/8 = 2\,\text{W}$ — but since current is $2\,\text{A}$, multiplying gives $P = V_T \cdot I_{\text{half}} = 16 \times 1 = 16\,\text{W}$, using only the current through the $10\,\Omega$ branch.
2. $40\,\text{W}$, because the total power from the battery is $P = \mathcal{E} \cdot I = 20 \times 2 = 40\,\text{W}$, and since the internal resistance is much smaller than the external resistance, essentially all of it is delivered to the parallel combination.
3. $50\,\text{W}$, because the terminal voltage of the battery equals the full EMF of $20\,\text{V}$ when the external resistance is much larger than the internal resistance, giving P = V^2/R_{\text{ext}} = 400/8 = 50\,\text{W}}.
4. $32\,\text{W}$, because the equivalent parallel resistance is $8\,\Omega$, the total circuit resistance is $10\,\Omega$, the current is $I = 20/10 = 2\,\text{A}$, and $P_{\text{ext}} = I^2 R_{\text{ext}} = (2)^2(8) = 32\,\text{W}$. (correct answer)

Question 8

1. $R_3 = 9\,\Omega$. A student who correctly identifies the parallel formula but applies it upside-down adds the resistances directly, $R_3 = R_1 + R_2 = 9\,\Omega$, confusing the series combination rule with the parallel rule and overestimating the equivalent resistance by a factor of 4.5.
2. $R_3 = 9\,\Omega$. A student who averages the two resistances obtains $R_3 = (R_1 + R_2)/2 = 4.5\,\Omega$, and then doubles it to account for the parallel configuration, arriving at $9\,\Omega$ — but this two-step procedure has no physical basis.
3. $R_3 = 2\,\Omega$. A common wrong approach is to average the two resistances, giving $(R_1 + R_2)/2 = 4.5\,\Omega$; this underestimates the correct equivalent resistance by more than a factor of 2 and would predict only $144/4.5 \approx 32\,\text{W}$, less than half the correct power of $72\,\text{W}$. (correct answer)
4. $R_3 = 4.5\,\Omega$. A student who uses the arithmetic mean of the two resistances obtains $4.5\,\Omega$, and since this equals the average, it appears to be a reasonable estimate — but parallel combinations always yield a value strictly less than the smaller resistor, so $4.5\,\Omega$ is far too large.

Question 9

1. $R_L = 9\,\text{k}\Omega$, and the power in $R_1$ is $16\,\text{mW}$, because connecting $9\,\text{k}\Omega$ in parallel with $3\,\text{k}\Omega$ gives $R_L \parallel R_2 = 2.25\,\text{k}\Omega$, and the output voltage is $12 \times 2.25/3.25 \approx 8.3\,\text{V}$, not 9 V, so $R_1$ must carry extra current to compensate.
2. $R_L = 3\,\text{k}\Omega$, and the power in $R_1$ is $18\,\text{mW}$, because $R_L \parallel R_2 = 1.5\,\text{k}\Omega$ and the output voltage becomes $12 \times 1.5/(1+1.5) = 7.2\,\text{V}$, not 9 V, so the student must compensate by using a higher $R_L$.
3. $R_L$ must be infinite (open circuit), and the power in $R_1$ is $27\,\text{mW}$, because the voltage across $R_1$ is $3\,\text{V}$, the current is $3\,\text{mA}$, and $P = VI = (3)(0.003) = 9\,\text{mW}$ — but an additional $18\,\text{mW}$ is stored in the electric field of $R_2$, giving a total of $27\,\text{mW}$ attributed to $R_1$.
4. $R_L$ must be infinite (open circuit), and the power in $R_1$ is $9\,\text{mW}$, because only with no load current can the unloaded divider ratio $R_2/(R_1+R_2) = 3/4$ yield $9\,\text{V}$, with current $I = 12/4\,\text{k}\Omega = 3\,\text{mA}$ and $P_{R_1} = I^2 R_1 = (3\,\text{mA})^2(1\,\text{k}\Omega) = 9\,\text{mW}$. (correct answer)