Electro Magnetic Induction And Alternating Currents Question 201
Question: In the circuit shown below, the key K is closed at t=0. The current through the battery is
Options:
A) $ \frac{VR_{1}R_{2}}{\sqrt{R_{1}^{2}+R_{2}^{2}}} $ at t=0 and $ \frac{V}{R_{2}} $ at $ t=\infty $
B) $ \frac{V}{R_{2}} $ at t=0 and $ \frac{V(R_{1}+R_{2})}{R_{1}R_{2}} $ at $ t=\infty $
C) $ \frac{V}{R_{2}} $ at t=0 and $ \frac{VR_{1}R_{2}}{\sqrt{R_{1}^{2}+R_{2}^{2}}} $ at $ t=\infty $
D) $ \frac{V(R_{1}+R_{2})}{R_{1}R_{2}} $ at t=0 and $ \frac{V}{R_{2}} $ at $ t=\infty $
Show Answer
Answer:
Correct Answer: C
Solution:
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At t = 0, no current will flow through L and $ R_{1} $
$ \therefore $ Current through battery $ =,\frac{V}{R_{2}} $
At $ t=\infty $ effective resistance, $ R_{eff}=\frac{R_{1}R_{2}}{R_{1}+R_{2}} $
$ \therefore $ Current through battery $ =\frac{V}{R_{eff}}=\frac{V,(R_{1}R_{2})}{(R_{1}+R_{2})} $
