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  • A current of I = 10 A is passed through the part of a circuit shown in the figure. What will be the potential difference between A and B when I is decreased at constant rate of <img alt="10^2" src=

A current of I = 10 A is passed through the part of a circuit shown in the figure. What will be the potential difference between A and B when I is decreased at constant rate of 10^2 amp/s, at the beginning?

Option: 1

8.5V


Option: 2

-8.5V


Option: 3

7.5V


Option: 4

-7.5V


Answers (1)

best_answer

Applying the law of potential between the points A and B we obtain,

\mathrm{\begin{aligned} & V_B-V_A=-I R+E-L \frac{d i}{d t} . \\ & \Rightarrow V_B-V_A=-10 \times 2+12-5 \times 10^{-3} \times 10^2 \\ & \Rightarrow V_B-V_A=-20+12-0.5 \\ & \Rightarrow V_B-V_A=-8.5 \text { volt. } \end{aligned}}

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Anam Khan

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