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(C) Applying Kirchhoff's voltage law from point $A$ to $B$:$V_A - iR - E - L\frac{di}{dt} = V_B$Given:$i = 5\, A$$R = 1\, \Omega$$E = 15\, V$$L = 5\,

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(C) Applying Kirchhoff's voltage law from point $A$ to $B$:$V_A - iR - E - L\frac{di}{dt} = V_B$Given:$i = 5\, A$$R = 1\, \Omega$$E = 15\, V$$L = 5\, mH = 5 	imes 10^{-3}\, H$$\frac{di}{dt} = -10^3\, A/s$ (since the current is decreasing)Substituting these values:$V_A - V_B = iR + E + L\frac{di}{dt}$$V_A - V_B = (5 	imes 1) + 15 + (5 	imes 10^{-3}) 	imes (-10^3)$$V_A - V_B = 5 + 15 - 5$$V_A - V_B = 15\, V$

The network shown in the figure is a part of a complete circuit. If at a certain instant the current $i$ is $5\, A$ and is decreasing at the rate of $10^3\, A/s$,then $V_A - V_B$ is......$V$.

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