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Transistor as an Amplifier

In the circui...

Question

In the circuit shown in figure, $E = 120 V, R_{1} = 3.0 Ω, R_{2} = 5.0 Ω$ and $L = 0.200 H$ . Switch $S$ is closed at $t = 0$ . The switch is closed for a long time and then opened. Just after the switch is opened. The potential difference $V_{c d}$ across the inductor $L$ is given by $- 48 x \times 10^{- 1} v o l t$ , then find $x$ .

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Solution

when $t = 0$ (refer fig.)
Ater long time,(refer fig.)
$I = \frac{E}{\frac{R_{1} R_{2}}{R_{1} + R_{2}}}$

$I = \frac{E (R_{1} + R_{2})}{R_{1} R_{2}}$

$I_{2} = \frac{I R_{1}}{R_{1} + R_{2}} = \frac{E}{R_{2}}$

When switch open,(refer fig.)

$V_{e d} - I_{2} R_{2} - I_{2} R_{1} = 0$

$V_{e d} = I_{2} (R_{1} + R_{2})$ = $\frac{E}{R_{2}} \times (R_{1} + R_{2})$
= $192$

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