In the following circuit,the switch S is clos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The circuit consists of two capacitors $C_1$ and $C_2$ connected in series with a resistor $R$ and a battery of $EMF$ $E$. When the switch $S$ is

Vedclass · Accepted Answer

${C_{eq}}E\,[1 - \exp ( - t/R{C_{eq}})]$

Vedclass · Answer

(A) The circuit consists of two capacitors $C_1$ and $C_2$ connected in series with a resistor $R$ and a battery of $EMF$ $E$. When the switch $S$ is closed at $t = 0$,the capacitors begin to charge. The equivalent capacitance of the series combination is given as $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$. The charging equation for a series $RC$ circuit is $Q(t) = Q_0(1 - e^{-t/	au})$,where $Q_0$ is the maximum charge and $	au = RC_{eq}$ is the time constant. The maximum charge on the equivalent capacitor is $Q_0 = C_{eq}E$. Since the capacitors are in series,the charge on each capacitor is the same and equal to the charge on the equivalent capacitor. Thus,the charge on $C_1$ as a function of time is $Q(t) = C_{eq}E[1 - \exp(-t/RC_{eq})]$.

In the following circuit,the switch $S$ is closed at $t = 0.$ The charge on the capacitor $C_1$ as a function of time will be given by $\left( {{C_{eq}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$

Similar Questions

Find the time constant for the given circuit.

How much time does it take for the current to decay to half of its initial value in the following charging circuit?

$A$ $20 \mu F$ capacitor is connected to a $45 \text{ V}$ battery through a circuit whose resistance is $2000 \Omega$. What is the final charge on the capacitor?

Vedclass Test Series

Exam Paper Generator

Online Exam Module