The circuit above works as a $12 V$ dc regulated voltage source. The power dissipated (in $mW$ ) in each diode is (considering both Zener diodes as identical and operating in breakdown mode)

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Solution

Current through the diodes $I = \frac{20 V - 12 V}{300 Ω + 500 Ω} = 10 mA$
Potential drop through each diode is $V = \frac{12}{2} = 6 V$
Thus, power dissipated in each diode is $P = V I = 6 \times 10 = 60 mA$

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The circuit above works as a 12 V dc regulated voltage source. The power dissipated (in mW) in each diode is (considering both Zener diodes as identical and operating in breakdown mode)

Current through the diodes I=20 V−12 V300 Ω+500 Ω=10 mA Potential drop through each diode is V=122=6 V Thus, power dissipated in each diode is P=VI=6×10=60 mA

The circuit above works as a $12 V$ dc regulated voltage source. The power dissipated (in $mW$ ) in each diode is (considering both Zener diodes as identical and operating in breakdown mode)

Current through the diodes $I = \frac{20 V - 12 V}{300 Ω + 500 Ω} = 10 mA$
Potential drop through each diode is $V = \frac{12}{2} = 6 V$
Thus, power dissipated in each diode is $P = V I = 6 \times 10 = 60 mA$