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(B) The current through the load resistor $R_L$ is given by:$i = \frac{V_z}{R_L} = \frac{10 \, V}{5 \, k\Omega} = 2 \, mA$The total current $I$ flowin

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$120$

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(B) The current through the load resistor $R_L$ is given by:$i = \frac{V_z}{R_L} = \frac{10 \, V}{5 \, k\Omega} = 2 \, mA$The total current $I$ flowing through the series resistor is:$I = \frac{V_{in} - V_z}{R_s} = \frac{24 \, V - 10 \, V}{1 \, k\Omega} = \frac{14 \, V}{1 \, k\Omega} = 14 \, mA$The current through the zener diode $I_z$ is:$I_z = I - i = 14 \, mA - 2 \, mA = 12 \, mA$The power dissipated across the zener diode is:$P = I_z 	imes V_z = 12 \, mA 	imes 10 \, V = 120 \, mW$

For the given circuit,the power across the zener diode is ..... $mW$.

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$A$ Zener diode of power rating $2 \, W$ is to be used as a voltage regulator. If the Zener diode has a breakdown voltage of $10 \, V$ and it has to regulate an unregulated input voltage that fluctuates between $6 \, V$ and $14 \, V$, the value of $R_s$ for safe operation should be $.... \Omega$.

Choose the correct statement about $Zener$ diode.

In the circuit given,the current through the Zener diode is ....... $mA$.

The value of the resistor,$R_S$,needed in the $dc$ voltage regulator circuit shown here,equals

The Zener diode is used for

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