Question

At what time has the current in the 8 Ω resistor decayed to half the value it had immediately after the switch was closed?(answer in ms)

The capacitors in the figure are charged and the switch closes at t=0s.

Answer 1

Assume of that the \(20 \mu \mathrm{F}\) capacitor as \(\mathrm{C}_{1}, 60 \mu \mathrm{F}\) as \(\mathrm{C}_{2}\) an \(20 \mathrm{C}_{3}\). Assume \(8 \Omega\) resistor as \(R_{1}, 30 \Omega\) resistor as \(R_{2}\), and \(20 \Omega\) resistor as \(R_{3}\). The equivalance resistance \(\left(R_{\text {eqv }}\right)=R_{1}+\left(\frac{1}{R_{2}}+\frac{1}{R_{3}}\right)\)

$$ \begin{array}{l} =8 \Omega+\left(\frac{1}{30 \Omega}+\frac{1}{20 \Omega}\right)^{-1} \\ =20 \Omega \end{array} $$

The equivalance capacitance \(\left(\mathrm{C}_{\text {eqv }}\right)=\frac{60 \mu \mathrm{F}}{2}+20 \mu \mathrm{F}\) \(=50 \mu \mathrm{F}\)

Here, \(I=\frac{1}{2} I_{0}\)

\(\frac{I}{I_{0}}=\frac{1}{2}=e^{-\frac{t}{R C}}\)

Here, \(R C=R_{\text {eqv }} C_{e q v}=20 \Omega(50 \mu \mathrm{F})=0.001 \mathrm{~s}\)

\(\ln \left(\frac{1}{2}\right)=-\frac{t}{R C}\)

\(t=-R C \ln \left(\frac{1}{2}\right)\)

\(=0.001 \mathrm{~s}(-0.693)\)

\(=6.93 \times 10^{-4} \mathrm{~s}\)

Hence, the time required for the current in the resistor to decayed to half the value it had is \(6.93 \times 10^{-4} \mathrm{~s}\)