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Standard XII

Physics

Kirchhoff's Junction Law

A part of a c...

Question

A part of a complete circuit is shown in the figure. At some instant, the value of current I is $1 A$ and it is decreasing at a rate of $10^{2} {As}^{- 1}$ . The value of the potential difference $(V_{P} - V_{Q})$ ( in volts ) at that instant, is

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Solution

Here, $L = 50 mH = 50 \times 10^{- 3} H; I = 1 A, R = 2 Ω$
Using potential drop concept, we can write
$V_{P} - L \frac{d I}{d t} - 30 + R I = V_{Q}$
$\Rightarrow V_{P} - V_{Q} = 50 \times 10^{- 3} \times 10^{2} + 30 - 1 \times 2$
$\Rightarrow V_{P} - V_{Q} = 5 + 30 - 2 = 33 V$

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A part of a complete circuit is shown in the figure. At some instant, the value of current I is 1 A and it is decreasing at a rate of 102 As−1. The value of the potential difference (VP−VQ) ( in volts ) at that instant, is

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A part of a complete circuit is shown in the figure. At some instant, the value of current I is $1 A$ and it is decreasing at a rate of $10^{2} {As}^{- 1}$ . The value of the potential difference $(V_{P} - V_{Q})$ ( in volts ) at that instant, is

Here, $L = 50 mH = 50 \times 10^{- 3} H; I = 1 A, R = 2 Ω$
Using potential drop concept, we can write
$V_{P} - L \frac{d I}{d t} - 30 + R I = V_{Q}$
$\Rightarrow V_{P} - V_{Q} = 50 \times 10^{- 3} \times 10^{2} + 30 - 1 \times 2$
$\Rightarrow V_{P} - V_{Q} = 5 + 30 - 2 = 33 V$