JEE MAIN - Physics (2017 (Offline) - No. 9)
In the given circuit diagram when the current reaches steady state in the
circuit, the charge on the capacitor of capacitance C will be:
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$$CE{{{r_1}} \over {({r_1} + r)}}$$
CE
$$CE{{{r_1}} \over {({r_2} + r)}}$$
$$CE{{{r_2}} \over {(r + {r_2})}}$$
Explanation
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
Vc = $${Q \over C}$$ = E - ir = E - $$\left( {{E \over {r + {r_2}}}} \right)r$$
$$ \therefore $$ Q = $$CE{{{r_2}} \over {(r + {r_2})}}$$
Current through the circuit,
i = $${E \over {r + {r_2}}}$$
Potential difference through capacitor
Vc = $${Q \over C}$$ = E - ir = E - $$\left( {{E \over {r + {r_2}}}} \right)r$$
$$ \therefore $$ Q = $$CE{{{r_2}} \over {(r + {r_2})}}$$
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