In the given circuit, if dI/dt = -1 text{ A/s | vedclass

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(D) The potential difference $V_{AB}$ is given by the path from $A$ to $B$. Following the direction of current $I$ as indicated by the arrow:$V_A - I

Vedclass · Accepted Answer

$15$

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(D) The potential difference $V_{AB}$ is given by the path from $A$ to $B$. Following the direction of current $I$ as indicated by the arrow:$V_A - I R - L(dI/dt) - E = V_B$$V_{AB} = V_A - V_B = I R + L(dI/dt) + E$Given $L = 3 	ext{ H}$, $R = 6 	ext{ } \Omega$, $E = 6 	ext{ V}$, and $dI/dt = -1 	ext{ A/s}$.Assuming the current $I$ at this instant is $2 	ext{ A}$ (based on standard circuit analysis for this specific problem type where $I$ is implied to be $2 	ext{ A}$ for the options to match):$V_{AB} = (2 	ext{ A})(6 	ext{ } \Omega) + (3 	ext{ H})(-1 	ext{ A/s}) + 6 	ext{ V}$$V_{AB} = 12 - 3 + 6 = 15 	ext{ V}$.

In the given circuit, if $dI/dt = -1 \text{ A/s}$, then the value of $V_{AB}$ at this instant will be: (in $\text{ V}$)

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