For the circuit shown in the figure,V_{AB} = | vedclass

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(A) The diode $D$ is in forward bias,so we can replace it with a short circuit (assuming an ideal diode with zero resistance). Now,the $10 \ \Omega$ r

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$10$

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(A) The diode $D$ is in forward bias,so we can replace it with a short circuit (assuming an ideal diode with zero resistance). Now,the $10 \ \Omega$ resistor in the middle and the $10 \ \Omega$ resistor on the right are in parallel. The equivalent resistance $R'$ of these two parallel resistors is: $R' = \frac{10 	imes 10}{10 + 10} = 5 \ \Omega$. Now,the circuit simplifies to a series circuit with a $30 \ V$ source,a $10 \ \Omega$ resistor $(R_1)$,and the equivalent $5 \ \Omega$ resistor $(R')$. Using the voltage divider rule,the voltage $V_{AB}$ across the parallel combination is: $V_{AB} = V 	imes \frac{R'}{R_1 + R'} = 30 	imes \frac{5}{10 + 5} = 30 	imes \frac{5}{15} = 10 \ V$.

For the circuit shown in the figure,$V_{AB} = $ ....... $V$.

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