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Standard X

Physics

Electric Potential and Potential Difference

The network s...

Question

The network shown in the figure is part of a complete circuit. What is the potential difference $V_{B} - V_{A}$ , when the current I is $5$ A and is decreasing at a rate of $10^{3}$ A/s?

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Solution

In accordance with law of potential distribution, for the given network,
$V_{A} - I R + E - L \frac{d I}{d t} = V_{B}$
And as I is decreasing here, $(d I / d t)$ is negative.
$V_{B} - V_{A} = - 5 \times 1 + 15 - 5 \times 10^{- 3} (- 10^{3})$
$V_{B} - V_{A} = - 5 + 15 + 5 = 15$ V.

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The network shown in the figure is part of a complete circuit. What is the potential difference VB−VA, when the current I is 5 A and is decreasing at a rate of 103 A/s?

In accordance with law of potential distribution, for the given network,VA−IR+E−LdIdt=VBAnd as I is decreasing here, (dI/dt) is negative.VB−VA=−5×1+15−5×10−3(−103)VB−VA=−5+15+5=15V.

The network shown in the figure is part of a complete circuit. What is the potential difference $V_{B} - V_{A}$ , when the current I is $5$ A and is decreasing at a rate of $10^{3}$ A/s?

In accordance with law of potential distribution, for the given network,
$V_{A} - I R + E - L \frac{d I}{d t} = V_{B}$
And as I is decreasing here, $(d I / d t)$ is negative.
$V_{B} - V_{A} = - 5 \times 1 + 15 - 5 \times 10^{- 3} (- 10^{3})$
$V_{B} - V_{A} = - 5 + 15 + 5 = 15$ V.