In the given circuit, if frac{dI}{dt} = -1 , | vedclass

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(A) To find the potential difference $(V_A - V_B)$, we apply Kirchhoff's Voltage Law $(KVL)$ starting from point $A$ to point $B$ along the circuit pa

Vedclass · Accepted Answer

$30$

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(A) To find the potential difference $(V_A - V_B)$, we apply Kirchhoff's Voltage Law $(KVL)$ starting from point $A$ to point $B$ along the circuit path.Starting from $A$, we move through the resistor $(12 \, \Omega)$, the inductor $(6 \, H)$, and the battery $(12 \, V)$ to reach $B$.The current $I = 2 \, A$ flows from $A$ to $B$.The potential drop across the resistor is $V_R = I \cdot R = 2 \, A 	imes 12 \, \Omega = 24 \, V$.The potential drop across the inductor is $V_L = L \frac{dI}{dt} = 6 \, H 	imes (-1 \, A/s) = -6 \, V$.The potential drop across the battery is $12 \, V$ (moving from positive to negative terminal).Applying $KVL$: $V_A - I \cdot R - L \frac{dI}{dt} - 12 = V_B$.Rearranging gives: $V_A - V_B = I \cdot R + L \frac{dI}{dt} + 12$.Substituting the values: $V_A - V_B = (2)(12) + 6(-1) + 12$.$V_A - V_B = 24 - 6 + 12 = 30 \, V$.

In the given circuit, if $\frac{dI}{dt} = -1 \, A/s$, then the value of $(V_A - V_B)$ at this instance will be (in $\, V$)

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