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(C) The potential at $A$ is $V_A = -10 	ext{ V}$ and at $B$ is $V_B = -2 	ext{ V}$.For the upper diode,the anode is at $V_A = -10 	ext{ V}$ and the

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$16$

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(C) The potential at $A$ is $V_A = -10 	ext{ V}$ and at $B$ is $V_B = -2 	ext{ V}$.For the upper diode,the anode is at $V_A = -10 	ext{ V}$ and the cathode is connected to the node between the $4 \Omega$ and $6 \Omega$ resistors. Since $V_A For the lower diode,the anode is connected to the node between the $8 \Omega$ and $2 \Omega$ resistors and the cathode is connected to the $12 \Omega$ resistor leading to $B$. Since $V_A When a diode is reverse-biased,it acts as an open circuit.Thus,the circuit simplifies to a series combination of the $8 \Omega$,$2 \Omega$,and $6 \Omega$ resistors.The equivalent resistance is $R_{eq} = 8 \Omega + 2 \Omega + 6 \Omega = 16 \Omega$.

In the following circuit,the equivalent resistance between $A$ and $B$ is ..... $\Omega$.

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