Compound Direct Current Circuits - AP Physics C Electricity and Magnetism

If a wire of negligible resistance is connected across the two terminals of the series resistor, effectively shorting it, what happens to the total power delivered by the battery?
Answer: It increases, because the total equivalent resistance of the circuit decreases. Let each resistor have resistance $R$ . The initial equivalent resistance is $R_{eq,1} = R + (1/R + 1/R)^{-1} = R + R/2 = 1.5R$ . When the series resistor is shorted, it is bypassed, and the new equivalent resistance is just that of the parallel pair, $R_{eq,2} = R/2$ . Since the total equivalent resistance decreases, the total current from the battery ( $I=V/R_{eq}$ ) increases. The total power delivered by the battery is $P = V^2/R_{eq}$ . Since $R_{eq}$ decreases, the total power $P$ increases.
Which of the following correctly compares the power $P_1$ , $P_2$ , and $P_3$ dissipated by the respective resistors?
Answer: $P_1 > P_3 > P_2$ Let the total current be $I$ . Power in $R_1$ is $P_1 = I^2 R_1 = 4I^2$ . The equivalent resistance of the parallel part is $R_p = (1/12 + 1/6)^{-1} = 4\,\Omega$ . The voltage across the parallel part is $V_p = I R_p = 4I$ . The power in $R_2$ is $P_2 = V_p^2 / R_2 = (4I)^2 / 12 = 16I^2/12 = (4/3)I^2$ . The power in $R_3$ is $P_3 = V_p^2 / R_3 = (4I)^2 / 6 = 16I^2/6 = (8/3)I^2$ . Comparing the powers: $P_1 = 4I^2$ , $P_2 \approx 1.33I^2$ , $P_3 \approx 2.67I^2$ . Therefore, the correct ranking is $P_1 > P_3 > P_2$ .
If the ideal voltmeter is replaced by a real voltmeter with a large but finite resistance, how will the measured potential difference $V_{real}$ compare to $V_{ideal}$ ?
Answer: $V_{real} < V_{ideal}$ because the voltmeter reduces the equivalent resistance of the part of the circuit it is connected to. A real voltmeter is connected in parallel with resistor $R_1$ . This creates a parallel combination with an equivalent resistance $R_{p1}$ that is less than $R_1$ . This decrease in resistance for one part of the series circuit leads to a decrease in the total equivalent resistance of the entire circuit. According to the voltage divider principle, the fraction of the total voltage that drops across the $R_1$ -voltmeter combination will be smaller than the fraction that dropped across $R_1$ alone. Thus, $V_{real} < V_{ideal}$ .
If the resistance of $R_2$ is decreased, what is the effect on the potential difference across $R_3$ ?
Answer: It decreases, because the total resistance of the circuit decreases. Decreasing $R_2$ decreases the equivalent resistance of the parallel combination ( $R_p$ ). This, in turn, decreases the total equivalent resistance of the circuit ( $R_{eq} = R_1 + R_p$ ). A lower total resistance causes the total current from the battery ( $I_{total}$ ) to increase. The potential difference across $R_1$ is $V_1 = I_{total}R_1$ , so $V_1$ increases. The potential difference across the parallel part (and thus across $R_3$ ) is given by Kirchhoff's loop rule: $V_p = V_3 = \mathcal{E} - V_1$ . Since $V_1$ increases, $V_3$ must decrease.
What is the equivalent resistance of this entire network of resistors?
Answer: $10.0\,\Omega$ First, find the resistance of each series branch. Branch A: $R_A = 3\,\Omega + 7\,\Omega = 10\,\Omega$ . Branch B: $R_B = 4\,\Omega + 6\,\Omega = 10\,\Omega$ . Next, find the equivalent resistance of these two branches in parallel: $R_{AB} = (1/10 + 1/10)^{-1} = (2/10)^{-1} = 5\,\Omega$ . Finally, this parallel combination is in series with the $5\,\Omega$ resistor, so the total equivalent resistance is $R_{eq} = R_{AB} + 5\,\Omega = 5\,\Omega + 5\,\Omega = 10.0\,\Omega$ .
Which of the following expressions correctly ranks the magnitudes of the currents?
Answer: $I_1 > I_2 > I_3$ The current $I_1$ flows through $R_1$ and is the total current for the circuit. This current then splits to flow through the parallel branches containing $R_2$ and $R_3$ . By Kirchhoff's junction rule, $I_1 = I_2 + I_3$ , so $I_1$ is the largest current. In a parallel combination, the potential difference across each branch is the same. Since $I = V/R$ , the branch with the lower resistance will have the higher current. Because $R_2 < R_3$ , it follows that $I_2 > I_3$ . Therefore, the correct ranking is $I_1 > I_2 > I_3$ .
What is the current flowing through the $3\,\Omega$ resistor?
Answer: $2.0\,\text{A}$ First, find the equivalent resistance of the parallel part: $R_p = (1/3 + 1/6)^{-1} = (3/6)^{-1} = 2\,\Omega$ . The total equivalent resistance is $R_{eq} = 2\,\Omega + R_p = 2\,\Omega + 2\,\Omega = 4\,\Omega$ . The total current from the battery is $I_{total} = V/R_{eq} = 12\,\text{V} / 4\,\Omega = 3\,\text{A}$ . The voltage across the parallel part is $V_p = I_{total} \times R_p = 3\,\text{A} \times 2\,\Omega = 6\,\text{V}$ . The current through the $3\,\Omega$ resistor is $I_3 = V_p / 3\,\Omega = 6\,\text{V} / 3\,\Omega = 2.0\,\text{A}$ .
A network of resistors with a total equivalent resistance $R_{ext}$ is connected to a battery with EMF $E$ and internal resistance $r$ . The current flowing is $I$ . If the same resistor network were connected to an ideal battery with the same EMF $E$ , the current would be $I_{ideal}$ . Which expression correctly relates $I_{ideal}$ to $I$ ?
Answer: $I_{ideal} = I(1 + r/R_{ext})$ For the nonideal battery, $I = \mathcal{E} / (R_{ext} + r)$ . For the ideal battery, $I_{ideal} = \mathcal{E} / R_{ext}$ . From the first equation, we can write $E = I(R_{ext} + r)$ . Substituting this into the second equation gives $I_{ideal} = I(R_{ext} + r) / R_{ext}$ . Distributing the denominator gives $I_{ideal} = I(R_{ext}/R_{ext} + r/R_{ext}) = I(1 + r/R_{ext})$ .
What is the effect of adding resistor $R_3$ on the total power dissipated by the circuit?
Answer: The total power decreases because the total equivalent resistance of the circuit increases. Adding a resistor $R_3$ in series increases the total equivalent resistance of the circuit. The total power dissipated by the circuit is given by $P = V^2 / R_{eq}$ , where $V$ is the constant voltage of the battery. Since $R_{eq}$ increases, the total power dissipated by the circuit must decrease.
If the ideal ammeter is replaced by a real ammeter with a nonzero internal resistance, how will the new current measurement compare to $I_0$ ?
Answer: It will be less than $I_0$ because the real ammeter increases the total resistance of the circuit. A real ammeter has a small but nonzero internal resistance. When placed in series in the circuit, this resistance adds to the existing resistance $R$ . This increases the total equivalent resistance of the circuit. According to Ohm's law, $I = V/R_{total}$ , an increase in the total resistance will cause a decrease in the total current. Therefore, the measured current will be less than the current measured by the ideal ammeter.
How does the addition of $R_3$ affect the potential difference across $R_1$ ?
Answer: It increases, because the total current from the battery increases. Adding $R_3$ in parallel with $R_2$ decreases the equivalent resistance of that branch. This, in turn, decreases the total equivalent resistance of the entire circuit. According to Ohm's law for the whole circuit, a lower total resistance results in a higher total current drawn from the battery. Since the potential difference across $R_1$ is given by $V_1 = I_{total}R_1$ , an increase in the total current leads to an increase in the potential difference across $R_1$ .
What is the potential difference across the parallel combination of the two $4\,\Omega$ resistors?
Answer: $6\,\text{V}$ The equivalent resistance of two $4\,\Omega$ resistors in parallel is $R_p = (1/4 + 1/4)^{-1} = 2\,\Omega$ . The total circuit resistance is the sum of the series resistor and this parallel equivalent: $R_{eq} = 2\,\Omega + 2\,\Omega = 4\,\Omega$ . The total current is $I = V/R_{eq} = 12\,\text{V} / 4\,\Omega = 3\,\text{A}$ . The potential difference across the parallel combination is $V_p = I \times R_p = 3\,\text{A} \times 2\,\Omega = 6\,\text{V}$ .
What is the approximate reading on the ideal ammeter?
Answer: $1.7\,\text{A}$ First, calculate the equivalent resistance of the parallel branch: $R_p = (1/12 + 1/6)^{-1} = (1/12 + 2/12)^{-1} = (3/12)^{-1} = 4\,\Omega$ . The total circuit resistance is $R_{eq} = 4\,\Omega + R_p = 4\,\Omega + 4\,\Omega = 8\,\Omega$ . The total current is $I_{total} = V/R_{eq} = 20\,\text{V}/8\,\Omega = 2.5\,\text{A}$ . The voltage across the parallel branch is $V_p = I_{total}R_p = 2.5\,\text{A} \times 4\,\Omega = 10\,\text{V}$ . The current through the $6\,\Omega$ resistor is $I_6 = V_p/6\,\Omega = 10\,\text{V}/6\,\Omega \approx 1.67\,\text{A}$ .
A resistor with resistance $R_1$ is connected in series with a parallel combination of two other resistors with resistances $R_2$ and $R_3$ . Which of the following expressions represents the equivalent resistance of this entire network?
Answer: $R_1 + \frac{R_2 R_3}{R_2 + R_3}$ The equivalent resistance of the parallel combination of $R_2$ and $R_3$ is $R_p = (\frac{1}{R_2} + \frac{1}{R_3})^{-1} = \frac{R_2 R_3}{R_2 + R_3}$ . This combination is in series with $R_1$ , so the total equivalent resistance is found by adding their resistances: $R_{eq} = R_1 + R_p = R_1 + \frac{R_2 R_3}{R_2 + R_3}$ .
If resistor $R_2$ is removed from the circuit, creating an open in its branch, what is the effect on the current passing through resistor $R_1$ ?
Answer: It decreases, because the total equivalent resistance of the circuit increases. Initially, the total resistance includes the parallel combination of $R_2$ and $R_3$ . The equivalent resistance of this parallel part is always less than either individual resistance. When $R_2$ is removed, the parallel combination is replaced by just $R_3$ . Since $R_3$ is greater than the equivalent resistance of the original parallel pair, the total equivalent resistance of the circuit increases. According to Ohm's Law ( $I=V/R_{eq}$ ), an increase in total resistance leads to a decrease in the total current. Since $R_1$ is in the main branch, the current through it is the total current, which decreases.
What is the terminal voltage of the battery when connected to this circuit?
Answer: $8.0\,\text{V}$ First, calculate the equivalent resistance of the external circuit. The parallel part is $R_p = (1/3 + 1/6)^{-1} = 2.0\,\Omega$ . The total external resistance is $R_{ext} = 2.0\,\Omega + R_p = 2.0\,\Omega + 2.0\,\Omega = 4.0\,\Omega$ . The total resistance of the entire circuit is $R_{total} = R_{ext} + r = 4.0\,\Omega + 0.5\,\Omega = 4.5\,\Omega$ . The current is $I = \mathcal{E} / R_{total} = 9.0\,\text{V} / 4.5\,\Omega = 2.0\,\text{A}$ . The terminal voltage is $V_T = \mathcal{E} - Ir = 9.0\,\text{V} - (2.0\,\text{A})(0.5\,\Omega) = 8.0\,\text{V}$ .
What is the ratio of the parallel current to the series current, $I_P / I_S$ ?
Answer: $16$ When the four resistors are in series, the equivalent resistance is $R_S = R+R+R+R = 4R$ . The current is $I_S = \mathcal{E}/(4R)$ . When the four resistors are in parallel, the equivalent resistance is $R_P = (1/R + 1/R + 1/R + 1/R)^{-1} = (4/R)^{-1} = R/4$ . The current is $I_P = \mathcal{E}/(R/4) = 4\mathcal{E}/R$ . The ratio is $I_P / I_S = (4\mathcal{E}/R) / (\mathcal{E}/(4R)) = 16$ .
An ideal voltmeter is connected in parallel with the $5\,\Omega$ resistor. If the voltmeter reads $15\,\text{V}$ , what is the EMF of the battery?
Answer: $37.5\,\text{V}$ The voltmeter reading gives the potential difference across the $5\,\Omega$ resistor. The current through this resistor (which is the total current) is $I = V/R = 15\,\text{V} / 5\,\Omega = 3\,\text{A}$ . Next, find the equivalent resistance of the parallel branch: $R_p = (1/10 + 1/30)^{-1} = (3/30 + 1/30)^{-1} = (4/30)^{-1} = 7.5\,\Omega$ . The potential difference across the parallel branch is $V_p = I R_p = 3\,\text{A} \times 7.5\,\Omega = 22.5\,\text{V}$ . The EMF of the battery is the sum of the potential differences across the series components: $EMF = V_5 + V_p = 15\,\text{V} + 22.5\,\text{V} = 37.5\,\text{V}$ .
A nonideal battery with EMF $E$ and internal resistance $r$ is connected to a variable external resistor $R$ . The power delivered to the external resistor is maximized when which of the following conditions is met?
Answer: The external resistance is equal to the internal resistance, $R=r$ . The current in the circuit is $I = \mathcal{E} / (R+r)$ . The power delivered to the external resistor is $P = I^2 R = (\frac{\mathcal{E}}{R+r})^2 R$ . To find the maximum power, we take the derivative of $P$ with respect to $R$ and set it to zero. The calculation shows that $dP/dR = 0$ when $R=r$ . This is the condition for maximum power transfer.
What is the resistance of $R_x$ ?
Answer: $6.0\,\Omega$ The total equivalent resistance of the circuit is $R_{eq} = V/I = 24\,\text{V} / 4.0\,\text{A} = 6.0\,\Omega$ . This total resistance is the sum of the series resistor and the equivalent resistance of the parallel part, $R_p$ . So, $R_{eq} = 3.0\,\Omega + R_p$ , which gives $R_p = 6.0\,\Omega - 3.0\,\Omega = 3.0\,\Omega$ . For the parallel combination, $1/R_p = 1/6.0 + 1/R_x$ . Substituting $R_p = 3.0\,\Omega$ , we get $1/3.0 = 1/6.0 + 1/R_x$ . This gives $1/R_x = 1/3.0 - 1/6.0 = 2/6.0 - 1/6.0 = 1/6.0$ . Therefore, $R_x = 6.0\,\Omega$ .

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