In the given circuit,calculate the potential | vedclass

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(C) Applying Kirchhoff's voltage law to the loop,we start from point $P$ and move clockwise:$V_P - I(1) - 8 + 12 - I(2) - I(9) = V_Q$$V_P - V_Q = I(1

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$3$

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(C) Applying Kirchhoff's voltage law to the loop,we start from point $P$ and move clockwise:$V_P - I(1) - 8 + 12 - I(2) - I(9) = V_Q$$V_P - V_Q = I(1 + 2 + 9) + 8 - 12$Since the circuit is a single loop,the current $I$ is determined by the net $EMF$ divided by total resistance:$I = \frac{12 - 8}{1 + 2 + 9} = \frac{4}{12} = \frac{1}{3} \, A$Now,calculate the potential difference $V_P - V_Q$:$V_P - V_Q = I(12) - 4 = \frac{1}{3}(12) - 4 = 4 - 4 = 0 \, V$Wait,re-evaluating the path from $P$ to $Q$ through the $9 \, \Omega$ resistor:$V_P - V_Q = I 	imes 9 = \frac{1}{3} 	imes 9 = 3 \, V$.

In the given circuit,calculate the potential difference between points $P$ and $Q$ in $volt$.

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