Question

A capacitor of capacitance 8.0 $\mu F$ is connected to a battery of emf 6.0 V through a resistance of 24 $\Omega$. Find the current in the circuit (a) just after the connections are made and (b) one time constant after the connections are made.

Answer 1

$C=8\mu f,E=6V,R=24\Omega$

$\left(a\right)i=\frac{V}{R}=\frac{6}{24}=0.25A$

$\left(b\right)q=Q{e}^{\frac{-t}{RC}}$

$=(8\times {10}^{-6}\times 6)e^{-1}$

$=48\times {10}^{-6}\times 0.63$

$=3.024\times {10}^{-5}$

$V=\frac{Q}{C}=\frac{3.024\times {10}^{-5}}{8\times {10}^{-5}}=3.78$

$E=V+iR$ $\Rightarrow 6=3.7+i24$

$\Rightarrow i=0.09A$