As shown in the figure,in steady state,the ch | vedclass

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

Question

(A) In steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit effectively cons

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) In steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit effectively consists of the battery $E = 10 \, V$ with internal resistance $r = 10 \, \Omega$ connected in series with the resistor $R = 100 \, \Omega$.The current $I$ in the circuit is given by $I = \frac{E}{R + r} = \frac{10}{100 + 10} = \frac{10}{110} = \frac{1}{11} \, A$.The potential difference $V_R$ across the resistor $R = 100 \, \Omega$ is $V_R = I 	imes R = \frac{1}{11} 	imes 100 = \frac{100}{11} \, V$.Since the capacitor is in parallel with the resistor $R$,the potential difference across the capacitor is equal to the potential difference across the resistor $R$.The charge $q$ stored in the capacitor is $q = C 	imes V_R = (1.1 	imes 10^{-6} \, F) 	imes \left(\frac{100}{11} \, Vight)$.$q = (1.1 	imes 10^{-6}) 	imes \left(\frac{100}{11}ight) = 0.1 	imes 10^{-6} 	imes 100 = 10 	imes 10^{-6} \, C$.Therefore,the charge stored is $10 	imes 10^{-6} \, C$.

As shown in the figure,in steady state,the charge stored in the capacitor is $....... \times 10^{-6} \, C$.

Similar Questions

In the circuit shown,the switch is shifted from position $1 \rightarrow 2$ at $t = 0$. The switch was initially in position $1$ for a long time. The graph between the charge on capacitor $C$ and time $t$ is:

The charge on the capacitor of capacitance $C$ shown in the figure below will be

Three identical capacitors are given a charge $Q$ each and they are then allowed to discharge through resistances $R_1, R_2,$ and $R_3$. Their charges,as a function of time,are shown in the graph below. The smallest of the three resistances is

The circuit shown here has two batteries of $8.0 \, V$ and $16.0 \, V$,three resistors of $3 \, \Omega$,$9 \, \Omega$,and $9 \, \Omega$,and a capacitor of $5.0 \, \mu F$. How much is the current $I$ in the circuit in steady state? ................... $A$

$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module