The network shown in the figure is a part of 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 $10^{3} A / s$ ?

Z e r o

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15 V

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20 V

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25 V

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Solution

The correct option is C $20 V$
Writing the voltage equation starting from point A
$V_{A} - i R_{1} + E - L \frac{d i}{d t} = V_{B}$
$V_{B} - V_{A} = - i R_{1} + E - L \frac{d i}{d t}$
$V_{B} - V_{A} = - 5 (1) + 15 - (5 \times 10^{- 3}) \times (- 10^{3})$
$V_{B} - V_{A} = 15 V$

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Similar questions

The figure shows a part of a complete circuit. The potential difference $V_{B} - V_{A}$ when the current I is 5A and is decreasing at a rate of $10^{3}$ A $s^{- 1}$ is given by

Q. The network shown in the figure is part of a complete circuit. If at a certain instant, the current

I

5 A

and it is decreasing at a rate of

10^{3} A s^{- 1}

then

V_{B} - V_{A}

equals :

The network shown in the figure is part of a complete circuit. If at a certain instant, the current I is 5 A and it is decreasing at a rate of $10^{3} A s^{- 1}$ then $V_{B} - V_{A}$ equals