In the circuit of given figure, the final voltage drop across the capacitor C is

Question

In the circuit of given figure, the final voltage drop across the capacitor C is (A)  (Vr1/r1+r2)  (B)  (Vr2/r1+r2)  (C)  (V(r1+r2)/r2)  (D)  (V(r1+r2)/r1+r2+r3)

Tardigrade Admin · Accepted Answer

In steady state, the capacitor arm produce an infinite resistance. So, the potential difference across

C

is that across

r_{2}

.
So, current through

r_{2} = \frac{V}{( r _{1} + r _{2} )}

Potential difference across

r_{2} = \frac{V ( r _{2} )}{( r _{1} + r _{2} )}

Q. In the circuit of given figure, the final voltage drop across the capacitor $C$ is

$\frac{V r _{1}}{r _{1} + r _{2}}$

$\frac{V r _{2}}{r _{1} + r _{2}}$

$\frac{V ( r _{1} + r _{2} )}{r _{2}}$

$\frac{V ( r _{1} + r _{2} )}{r _{1} + r _{2} + r _{3}}$

In steady state, the capacitor arm produce an infinite resistance. So, the potential difference across $C$ is that across $r_{2}$ .
So, current through $r_{2} = \frac{V}{( r _{1} + r _{2} )}$
Potential difference across $r_{2} = \frac{V ( r _{2} )}{( r _{1} + r _{2} )}$

Q. In the circuit of given figure, the final voltage drop across the capacitor C is

r1​+r2​Vr1​​

r1​+r2​Vr2​​

r2​V(r1​+r2​)​

r1​+r2​+r3​V(r1​+r2​)​