Question

A capacitor of 30.0 μ F 30.0 μF and a resistor of 110 Ω 110 Ω are quickly connected in series to a battery of 4.00 V. 4.00 V. What is the charge Q Q on the capacitor 0.00100 s 0.00100 s after the connection is made?

Answer 1

Answer:$Q=31.368\ \mu C$

Explanation:

Given

capacitance $C=30\ muF$

Resistance $R=110\ Omega$

Applied Voltage $V=4\ V$

$t=0.001\ s$

Charge on Capacitor in a R-C circuit is given by

$Q=Q_0(1-e^{\frac{-t}{RC}})$

$Q=CV(1-e^{\frac{-t}{RC}})$

$Q=30\times 10^{-6}\times 4(1-e^{\frac{-0.001}{110\times 30\times 10^{-6}}})$

$Q=120\times 10^{-6}\times 0.2614$

$Q=31.368\ \mu C$