The circuit shown below is working as an 8; V | vedclass

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Question

(D) The total resistance in the circuit is $R = 200 \; \Omega + 200 \; \Omega = 400 \; \Omega$.The voltage drop across the resistors is $V_R = V_{in}

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) The total resistance in the circuit is $R = 200 \; \Omega + 200 \; \Omega = 400 \; \Omega$.The voltage drop across the resistors is $V_R = V_{in} - V_{out} = 12 \; V - 8 \; V = 4 \; V$.The current flowing through the circuit is $I = \frac{V_R}{R} = \frac{4 \; V}{400 \; \Omega} = 0.01 \; A = 10 \; mA$.Since the two identical Zener diodes are connected in series, the voltage across each diode is $V_d = \frac{8 \; V}{2} = 4 \; V$.The power dissipated in each diode is $P = V_d 	imes I = 4 \; V 	imes 10 \; mA = 40 \; mW$.

The circuit shown below is working as an $8\; V\; dc$ regulated voltage source. When $12 \; V$ is used as input, the power dissipated (in $mW$) in each diode is (considering both zener diodes are identical).

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