In the given figure,the charge stored in 6 m | vedclass

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

Question

(B) At steady state,capacitors behave as open circuits. When points $A$ and $B$ are connected by a wire,the capacitors are bypassed in the $DC$ circui

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) At steady state,capacitors behave as open circuits. When points $A$ and $B$ are connected by a wire,the capacitors are bypassed in the $DC$ circuit path.The circuit effectively consists of the $6 \ \Omega$ resistor and the $3 \ \Omega$ resistor in series connected to the $9 \ V$ source.The equivalent resistance is $R_{	ext{eq}} = 6 \ \Omega + 3 \ \Omega = 9 \ \Omega$.The current flowing through the circuit is $i = \frac{V}{R_{	ext{eq}}} = \frac{9 \ V}{9 \ \Omega} = 1 \ A$.Since points $A$ and $B$ are connected,they are at the same potential. The potential at $A$ (and $B$) relative to the ground is the potential drop across the $3 \ \Omega$ resistor: $V_B = i 	imes 3 \ \Omega = 1 \ A 	imes 3 \ \Omega = 3 \ V$.The potential at the top terminal is $9 \ V$. Thus,the potential difference across the $6 \ \mu F$ capacitor (connected between the $9 \ V$ terminal and point $B$) is $\Delta V = 9 \ V - 3 \ V = 6 \ V$.The charge stored in the $6 \ \mu F$ capacitor is $Q = C \Delta V = 6 \ \mu F 	imes 6 \ V = 36 \ \mu C$.

In the given figure,the charge stored in $6 \ \mu F$ capacitor,when points $A$ and $B$ are joined by a connecting wire is . . . . . . $\mu C$.

Similar Questions

For the given circuit,find the voltage across the $2\ \mu F$ capacitor. (in $V$)

$125$ identical drops,each charged to a potential of $50\;V$,are combined to form a single drop. The potential of the new drop will be......$V$.

Find the potentials at points $A$ and $B$ in the circuit shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module