In an LR circuit with L = 2 H and R = 4, find the time constant and the current at t = 0.5 s when a 12 V battery is — JEE Physics
In an LR circuit with $L = 2$ H and $R = 4\ \Omega$, find the time constant and the current at $t = 0.5$ s when a 12 V battery is switched in.
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2mo ago
▲ 6
$$\tau = \frac{L}{R} = \frac{2}{4} = 0.5 s$$
$$I(t) = \frac{\mathcal{E}}{R}\left(1-e^{-t/\tau}\right) = 3\left(1 - e^{-1}\right) = 3(1-0.368) = 1.9 A$$
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Discussion (3)
P
This is exactly the kind of step-by-step I needed. Respect.
D
Is there a faster shortcut for this in the actual exam? Time is tight.
NR
Adding for context: NCERT covers the base concept in the same chapter.