A capacitor of capacitance $10\, mF$ is connected to a battery of emf $2\,V.$ It is found that it takes $50\, ms$ for the charge on the capacitor to become $12.6 \,mC.$ Then the resistance of the circuit is :.......$\, k \Omega$ (Take $1/e = 0.37$):-
Medium
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Charge at steady steady state ${{\rm{q}}_0} = 20\mu {\rm{C}}$
${\rm{q}} = {{\rm{q}}_0}\left( {1 - {{\rm{e}}^{{\rm{t/}}\tau }}} \right)$
${\rm{q}} = 12.6 = 20\left( {1 - {{\rm{e}}^{ - 50 \times {{10}^{ - 3{\rm{/}}\tau }}}}} \right)$
and $\tau = RC = 50 \times {10^{ - 3}} \Rightarrow R = 5\,{\rm{K}}\Omega $
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