Unit 11: Electric Circuits
Weightage : 15–18%
The figure above shows a \(10\,\mathrm{V}\) battery connected in a circuit with two resistors, a parallel-plate capacitor of capacitance \(C\), and three ammeters. The circuit has been connected for a long time.
Let \(I_i\) be the current in ammeter \(A_i\). Which of the following correctly ranks the currents in the three ammeters?
Correct Answer: \( \boxed{\mathrm{A}} \)
After the circuit has been connected for a long time, the capacitor is fully charged. A fully charged capacitor behaves as an open circuit, so no steady current flows through the capacitor branch.
Therefore,
\( I_2=0 \).
The remaining current flows only through the outer loop containing the \(50\,\Omega\) and \(100\,\Omega\) resistors. Since these resistors are in series, the same current passes through every point in the loop.
Thus,
\( I_1=I_3 \).
Using Ohm’s law, the steady current in the outer loop is
\( I=\dfrac{V}{R_{\mathrm{eq}}}=\dfrac{10}{50+100}=\dfrac{10}{150}=\dfrac{1}{15}\,\mathrm{A}\approx0.067\,\mathrm{A}. \)
Hence,
\( I_1=I_3=\dfrac{1}{15}\,\mathrm{A}>I_2=0. \)
Therefore, the correct ranking is \((I_1=I_3)>I_2\), corresponding to (A).
In the circuit represented above, two resistors (\(R_1\) and \(R_2\)), a capacitor \(C\), and an open switch \(S\) are connected to a battery. The circuit reaches equilibrium. The switch is then closed, and the circuit is allowed to come to a new equilibrium.
Which of the following is a true statement about the energy stored in the capacitor after the switch is closed compared with the energy stored in the capacitor before the switch is closed?
Correct Answer: \( \boxed{\mathrm{B}} \)
Initially, with the switch open and after a long time, no current flows through the circuit. Therefore, there is no voltage drop across \(R_1\), and the capacitor charges to the full battery voltage.
The initial energy stored in the capacitor is
\( U_i=\dfrac{1}{2}CV_{\text{battery}}^2. \)
When the switch is closed and the circuit reaches a new equilibrium, current flows through both \(R_1\) and \(R_2\). As a result, there is a voltage drop across \(R_1\), so the potential difference across the capacitor is less than the battery voltage.
Since the energy stored in a capacitor is
\( U=\dfrac{1}{2}CV^2, \)
a smaller voltage across the capacitor means the stored energy decreases.
Equivalently, some charge leaves the capacitor through the new loop formed when \(R_2\) is connected, reducing both the capacitor’s charge and its voltage.
Therefore, the energy stored in the capacitor after the switch is closed is less than before the switch is closed.
Therefore, the correct answer is (B).
A \(25\,\Omega\) resistor and a \(100\,\Omega\) resistor are connected in a circuit with a third resistor of unknown resistance \(R\), a capacitor of unknown capacitance \(C\), and a battery of unknown emf \( \varepsilon \), as shown above.
After a long time, the potential difference across the \(25\,\Omega\) resistor is measured to be \(4\,\mathrm{V}\). What is the emf of the battery?
Correct Answer: \( \boxed{\mathrm{D}} \)
After a long time, the capacitor is fully charged and behaves as an open circuit. The capacitor branch carries no current, leaving the three resistors \(25\,\Omega\), \(100\,\Omega\), and \(R\) connected in series.
The current in the circuit can be found from the voltage across the \(25\,\Omega\) resistor:
\( I=\dfrac{V}{R}=\dfrac{4}{25}=0.16\,\mathrm{A}. \)
The battery emf is the sum of the voltage drops across all three resistors:
\( \varepsilon=I(25+100+R). \)
Substituting the known current gives
\( \varepsilon=0.16(125+R). \)
Since the value of \(R\) is unknown, the emf cannot be calculated uniquely.
Therefore, the correct answer is (D).
