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Byju's Answer

Standard XII

Physics

RLC Circuit

In an L-C-R c...

Question

In an L-C-R circuit as shown both switches are open initially. Now switch $S_{1}$ and $S_{2}$ , are closed. ( $q$ is the charge on the capacitor and $r = R C$ is capacitance time constant). What is the charge on the capacitor at $t = 2 τ$ ?

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Solution

When $s_{1}$ is connected, and $s_{2}$ is kept open, it becomes a RC circuit.
So, at any time t, charge on capacitor C is
q= $(1 - e^{\frac{- t}{τ}})$
At $t = 2 τ$
Charge in the capacitor, $q = C V (1 - e^{- 2}$ )

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In an L-C-R circuit as shown both switches are open initially. Now switch S1 and S2, are closed. (q is the charge on the capacitor and r=RC is capacitance time constant). What is the charge on the capacitor at t=2τ?

When s1 is connected, and s2 is kept open, it becomes a RC circuit.So, at any time t, charge on capacitor C isq= (1−e−tτ)At t=2τCharge in the capacitor, q=CV(1−e−2)

In an L-C-R circuit as shown both switches are open initially. Now switch $S_{1}$ and $S_{2}$ , are closed. ( $q$ is the charge on the capacitor and $r = R C$ is capacitance time constant). What is the charge on the capacitor at $t = 2 τ$ ?

When $s_{1}$ is connected, and $s_{2}$ is kept open, it becomes a RC circuit.
So, at any time t, charge on capacitor C is
q= $(1 - e^{\frac{- t}{τ}})$
At $t = 2 τ$
Charge in the capacitor, $q = C V (1 - e^{- 2}$ )