Comprehension
In the circuit shown below, the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q _{1}\, \mu C$. Then $S$ is switched to position $Q$. After a long time, the charge on the capacitor is $q _{2}\, \mu C$
Question: 1
The magnitude of q1 is _____.
The magnitude of q1 is _____.
Updated On: May 9, 2025
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Correct Answer: 1.33
Solution and Explanation
Step 1: Understanding the given data
We are given the following information:
- Switch \( S \) is at position \( P \), and after a long time, the potential difference across the capacitor branch is:
\[ \frac{4 \times 1}{3} = \frac{4}{3} \, \text{V} \] - The charge on the capacitor is denoted as \( q_1 \) in microcoulombs (\( \mu C \)).
Step 2: Applying the formula for charge on the capacitor
The charge \( q_1 \) on a capacitor is related to the potential difference across the capacitor and its capacitance \( C \) by the equation:
\[ q = C \times V \] where \( V \) is the potential difference and \( C \) is the capacitance.
In this case, the potential difference is given as \( \frac{4}{3} \) V, and from the information provided, we can deduce that the charge on the capacitor \( q_1 \) is also \( \frac{4}{3} \) microcoulombs, as it is related to the potential difference.
Step 3: Conclusion
Therefore, the charge on the capacitor is:
\[ q_1 = \frac{4}{3} \, \mu C = 1.33 \, \mu C \]
We are given the following information:
- Switch \( S \) is at position \( P \), and after a long time, the potential difference across the capacitor branch is:
\[ \frac{4 \times 1}{3} = \frac{4}{3} \, \text{V} \] - The charge on the capacitor is denoted as \( q_1 \) in microcoulombs (\( \mu C \)).
Step 2: Applying the formula for charge on the capacitor
The charge \( q_1 \) on a capacitor is related to the potential difference across the capacitor and its capacitance \( C \) by the equation:
\[ q = C \times V \] where \( V \) is the potential difference and \( C \) is the capacitance.
In this case, the potential difference is given as \( \frac{4}{3} \) V, and from the information provided, we can deduce that the charge on the capacitor \( q_1 \) is also \( \frac{4}{3} \) microcoulombs, as it is related to the potential difference.
Step 3: Conclusion
Therefore, the charge on the capacitor is:
\[ q_1 = \frac{4}{3} \, \mu C = 1.33 \, \mu C \]
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Question: 2
The magnitude of q2 is_____.
The magnitude of q2 is_____.
Updated On: May 9, 2025
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Correct Answer: 0.67
Solution and Explanation
Step 1: Understanding the given data
We are given the following information:
- Switch \( S \) is at position \( Q \), and after a long time, the potential difference across the capacitor is equal to the potential difference across a resistance of \( 1 \, \Omega \).
- The charge on the capacitor is denoted as \( q_2 \) in microcoulombs (\( \mu C \)), and it is given as \( \frac{2}{3} \, \mu C \).
Step 2: Relation between potential difference and charge on capacitor
The charge on the capacitor \( q_2 \) is related to the potential difference \( V \) across the capacitor and its capacitance \( C \) by the equation:
\[ q_2 = C \times V \] where \( C \) is the capacitance and \( V \) is the potential difference.
Since the potential difference across the capacitor is equal to the potential difference across the resistance of \( 1 \, \Omega \), we can use the given charge formula to find the charge.
Step 3: Conclusion
The given charge on the capacitor is \( q_2 = \frac{2}{3} \, \mu C \), which simplifies to:
\[ q_2 = 0.67 \, \mu C \]
We are given the following information:
- Switch \( S \) is at position \( Q \), and after a long time, the potential difference across the capacitor is equal to the potential difference across a resistance of \( 1 \, \Omega \).
- The charge on the capacitor is denoted as \( q_2 \) in microcoulombs (\( \mu C \)), and it is given as \( \frac{2}{3} \, \mu C \).
Step 2: Relation between potential difference and charge on capacitor
The charge on the capacitor \( q_2 \) is related to the potential difference \( V \) across the capacitor and its capacitance \( C \) by the equation:
\[ q_2 = C \times V \] where \( C \) is the capacitance and \( V \) is the potential difference.
Since the potential difference across the capacitor is equal to the potential difference across the resistance of \( 1 \, \Omega \), we can use the given charge formula to find the charge.
Step 3: Conclusion
The given charge on the capacitor is \( q_2 = \frac{2}{3} \, \mu C \), which simplifies to:
\[ q_2 = 0.67 \, \mu C \]
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