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(C) First,we check if the Zener diode is in breakdown by calculating the voltage across the parallel branch without the Zener diode: $V_{open} = \frac

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$10$

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(C) First,we check if the Zener diode is in breakdown by calculating the voltage across the parallel branch without the Zener diode: $V_{open} = \frac{400 \ \Omega}{100 \ \Omega + 400 \ \Omega} 	imes 12 \ V = \frac{400}{500} 	imes 12 \ V = 0.8 	imes 12 \ V = 9.6 \ V$. Since the calculated voltage $9.6 \ V$ is greater than the Zener breakdown voltage $V_z = 4 \ V$,the Zener diode operates in the breakdown region. Therefore,the voltage across the $400 \ \Omega$ resistor is fixed at the Zener voltage,$V = 4 \ V$. The current $I$ through the ammeter is given by Ohm's law: $I = \frac{V}{R} = \frac{4 \ V}{400 \ \Omega} = 0.01 \ A = 10 \ mA$.

In the following circuit,the reading of the ammeter will be (Take Zener breakdown voltage $= 4 \ V$) (in $mA$)

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