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When the current in the portion of the circuit shown in the figure is `2A` and increasing at the rate of `1A//s`,the measured potential difference `V_(a)-V_(b)=8V`.However when the current is `2A` and decreasing at the rate of `1A//s`, the measured potential difference `V_(a)-V_(b)=4V`.The values of `R` and `L` are: A. `3Omega` and 2H, respectivelyB. `2Omega` and 3H, respectivelyC. `3Omega` and 2H, respectivelyD. |
Answer» Correct Answer - A |
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