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Standard XII

Physics

Resistance of P-N Junction

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. The key $K$ is inserted at time $t = 0$ . The initial $(t = 0)$ and final $(t \to \infty)$ currents through 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 $\frac{1}{15} A, \frac{1}{10} A$
At $t = 0$ the inductor behaves like infinite resistance.
Thus, $i = \frac{2}{30} = \frac{1}{15} A$

At $t = \infty$ the inductor behaves like a conducting wire and $10 Ω$ is short circuited.
$i = \frac{2}{20} = \frac{1}{10} A$

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

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

The correct option is A $\frac{1}{15} A, \frac{1}{10} A$
At $t = 0$ the inductor behaves like infinite resistance.
Thus, $i = \frac{2}{30} = \frac{1}{15} A$

At $t = \infty$ the inductor behaves like a conducting wire and $10 Ω$ is short circuited.
$i = \frac{2}{20} = \frac{1}{10} A$