For the circuit shown below,the current throu | vedclass

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(A) The Zener diode is connected in parallel with the $10 \, k\Omega$ resistor. Since the Zener breakdown voltage is $50 \, V$,the voltage across the

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$9$

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(A) The Zener diode is connected in parallel with the $10 \, k\Omega$ resistor. Since the Zener breakdown voltage is $50 \, V$,the voltage across the $10 \, k\Omega$ resistor is also $50 \, V$.The current through the $10 \, k\Omega$ resistor is $I_L = \frac{V_Z}{R_L} = \frac{50 \, V}{10 \, k\Omega} = 5 \, mA$.The total current supplied by the $120 \, V$ source passes through the $5 \, k\Omega$ series resistor. The voltage drop across this resistor is $120 \, V - 50 \, V = 70 \, V$.The total current $I_S$ is $I_S = \frac{70 \, V}{5 \, k\Omega} = 14 \, mA$.Applying Kirchhoff's Current Law at the node,the current through the Zener diode $I_Z$ is $I_Z = I_S - I_L = 14 \, mA - 5 \, mA = 9 \, mA$.

For the circuit shown below,the current through the Zener diode is......$mA$

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