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Standard XII

Physics

Transformer

Two resistors...

Question

Two resistors of $10 Ω$ and $20 Ω$ and an ideal inductor of 10 H are connected to a 2 V battery as shown in Fig. Key K is inserted at time $t = 0$ . The initial $(t = 0)$ and final $(t \to \infty)$ currents through the battery are:

\frac{1}{15} A, \frac{1}{10} A

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\frac{1}{10} A, \frac{1}{15} A

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\frac{2}{15} A, \frac{1}{10} A

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\frac{1}{15} A, \frac{2}{25} A

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Solution

The correct option is A $\frac{1}{15} A, \frac{1}{10} A$
Initially, inductor does not allow flux through it to change, therefore current passes through the parallel resistor. Therefor, at t=0, current through battery $I = \frac{2}{(10 + 20)} = \frac{1}{15} A$
Now at t= infinity, inductor acts as a short.
Current through battery $I = \frac{2}{20} = \frac{1}{10} A$

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Two resistors of 10Ω and 20Ω and an ideal inductor of 10 H are connected to a 2 V battery as shown in Fig. Key K is inserted at time t=0. The initial (t=0) and final (t→∞) currents through the battery are:

The correct option is A 115A,110AInitially, inductor does not allow flux through it to change, therefore current passes through the parallel resistor. Therefor, at t=0, current through batteryI=2(10+20)=115ANow at t= infinity, inductor acts as a short. Current through battery I=220=110A

Two resistors of $10 Ω$ and $20 Ω$ and an ideal inductor of 10 H are connected to a 2 V battery as shown in Fig. Key K is inserted at time $t = 0$ . The initial $(t = 0)$ and final $(t \to \infty)$ currents through the battery are: