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Question

$(A)$ The Zener diode maintains a constant voltage $V_z = 9\,V$ across the load resistance $R_L = 450\,\Omega$.The current through the load resistance

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$60$

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$(A)$ The Zener diode maintains a constant voltage $V_z = 9\,V$ across the load resistance $R_L = 450\,\Omega$.The current through the load resistance is $I_L = \frac{V_z}{R_L} = \frac{9\,V}{450\,\Omega} = 0.02\,A$.The maximum current through the Zener diode is $I_{Z,max} = \frac{P_{max}}{V_z} = \frac{0.27\,W}{9\,V} = 0.03\,A$.The total current flowing through the series resistor $R_s$ is $I = I_L + I_{Z,max} = 0.02\,A + 0.03\,A = 0.05\,A$.The voltage drop across the series resistor $R_s$ is $V_{R_s} = V_{in} - V_z = 12\,V - 9\,V = 3\,V$.Therefore,the value of the series resistor is $R_s = \frac{V_{R_s}}{I} = \frac{3\,V}{0.05\,A} = 60\,\Omega$.

$A$ $9\,V$ stabilized voltage supply is required to run a car stereo system from the car's $12\,V$ battery. $A$ Zener diode with $V_z = 9\,V$ and $P_{max} = 0.27\,W$ is to be used as the voltage regulator. The load resistance is $450\,\Omega$. The value of the series resistor $R_s$ is....$\Omega$.

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