Sum
A 20 μF capacitor is joined to a battery of emf 6.0 V through a resistance of 100 Ω. Find the charge on the capacitor 2.0 ms after the connections are made.
Advertisement Remove all ads
Solution
The growth of charge across a capacitor,
\[Q = Q_0 . e^{- \frac{t}{RC}} \]
\[ Q_0 = CV = 20 \times 6 \times {10}^{- 6} C\]
\[ \Rightarrow Q = 120 \times {10}^{- 6} . e^{- \frac{2 \times {10}^{- 3}}{{10}^2 . 20 \times {10}^{- 6}}} \]
\[ \Rightarrow Q = 120 \times {10}^{- 6} . e^{- 1} = 76 \mu C\]
Concept: Energy Stored in a Capacitor
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads
