Question 374: Electrostatics & Capacitance

Question

In the given circuit diagram when the current reaches steady state in the
circuit, the charge on the capacitor of capacitance C will be:

Accepted Answer

In steady state, flow of current through capacitor will
be zero.
Current through the circuit,
i = $${E \over {r + {r_2}}}$$
Potential difference through capacitor
V
c
= $${Q \over C}$$ = E - ir = E - $$\left( {{E \over {r + {r_2}}}} ight)r$$
$$ 	herefore $$ Q = $$CE{{{r_2}} \over {(r + {r_2})}}$$

Answer

$$CE{{{r_1}} \over {({r_1} + r)}}$$

Answer

CE

Answer

$$CE{{{r_1}} \over {({r_2} + r)}}$$

Answer

$$CE{{{r_2}} \over {(r + {r_2})}}$$

Electrostatics & Capacitance

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be:

Choose the correct answer: