Question:
Three capacitors each of capacitance \(12\,\mu F\), are connected in series. When this combination is connected to a battery of \(12\,V\), the charge drawn from the battery is
Three capacitors each of capacitance \(12\,\mu F\), are connected in series. When this combination is connected to a battery of \(12\,V\), the charge drawn from the battery is
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In series: capacitance decreases.
Updated On: Apr 24, 2026
- \(32\,\mu C\)
- \(24\,\mu C\)
- \(48\,\mu C\)
- \(16\,\mu C\)
- \(12\,\mu C\)
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The Correct Option is C
Solution and Explanation
Concept:
For series capacitors:
\[
\frac{1}{C_{eq}} = \sum \frac{1}{C}
\]
Step 1: Equivalent capacitance.
\[ \frac{1}{C_{eq}} = \frac{3}{12} \Rightarrow C_{eq} = 4\,\mu F \]
Step 2: Charge.
\[ Q = CV = 4 \times 12 = 48\,\mu C \]
Step 1: Equivalent capacitance.
\[ \frac{1}{C_{eq}} = \frac{3}{12} \Rightarrow C_{eq} = 4\,\mu F \]
Step 2: Charge.
\[ Q = CV = 4 \times 12 = 48\,\mu C \]
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