Question

Consider the circuit shown in figure. When switch $S_1$ closed and the other two switches open, the circuit has a time constant $0.05\text{sec}$. When switch $S_2$ closed and the other two switches open, the circuit has a time consatnt $2\text{sec}$. When switch $S_3$ closed an the other two switches open,the circuit oscillates with a period $T$. Find $T(\text{in sec). Take}{\pi }^2=10$

Answer 1

When switch $S_1$ is closed and the others are open, the inductor is escentially out of the circuit and what remains is an RC circuit
The time consatnt is ${\tau }_C=RC$.
Given ${\tau }_C=0.05\text{sec}$
When switch $S_2$ is closed and the others are open, the capacitor is essentially out of the circuit, in this case, what we have an $LR$ circuit with the time constant ${\tau }_L=\frac{L}{R}$
Given ${\tau }_L=2\text{sec}$

Finally, when switch $S_3$ is closed and the others are open,the resistor is essentially out of then circuit and what remains is an $LC$circuit that oscillates with time period.
$\Rightarrow T=2\pi \sqrt{LC}$
Here $LC={\tau }_C{\tau }_L=0.05\times 2=0.1$
$\Rightarrow T=2\pi \sqrt{\frac{1}{10}}=2\text{sec}$