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(C) From the circuit diagram,diodes ${D}_{1}$ and ${D}_{2}$ are in forward bias,so each has a resistance of $30\, \Omega$. Diode ${D}_{3}$ is in rever

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$2$

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(C) From the circuit diagram,diodes ${D}_{1}$ and ${D}_{2}$ are in forward bias,so each has a resistance of $30\, \Omega$. Diode ${D}_{3}$ is in reverse bias,so its resistance is infinite.The total resistance of the two parallel branches containing ${D}_{1}$ and ${D}_{2}$ is:${R}_{p} = \frac{(30 + 130)}{2} = \frac{160}{2} = 80\, \Omega$.The total resistance of the circuit is ${R}_{total} = {R}_{p} + 20\, \Omega = 80\, \Omega + 20\, \Omega = 100\, \Omega$.Using Ohm's law,the total current ${I}_{1}$ is:${I}_{1} = \frac{V}{{R}_{total}} = \frac{200\, V}{100\, \Omega} = 2\, A$.

In the given figure,each diode has a forward bias resistance of $30\, \Omega$ and infinite resistance in reverse bias. The current ${I}_{1}$ will be $....\,A$.

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