Question 25

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

Correct Answer: 33

$$V_Q - IR + V_{\text{battery}} - L\frac{dI}{dt} = V_P$$

$$V_Q - (1)(2) + 30 - \left(50 \times 10^{-3}\right)\left(-10^2\right) = V_P$$

$$V_Q - 2 + 30 + 5 = V_P \implies V_Q + 33 = V_P \implies V_P - V_Q = 33\text{ V}$$

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