The value of power dissipated across the Zene | vedclass

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(A) $1$. The voltage across the series resistor $R_{S}$ is $V_{R_{S}} = V_{in} - V_{Z} = 22 	ext{ V} - 15 	ext{ V} = 7 	ext{ V}$.$2$. The total cur

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$5$

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(A) $1$. The voltage across the series resistor $R_{S}$ is $V_{R_{S}} = V_{in} - V_{Z} = 22 	ext{ V} - 15 	ext{ V} = 7 	ext{ V}$.$2$. The total current flowing through the series resistor $R_{S}$ is $I = \frac{V_{R_{S}}}{R_{S}} = \frac{7 	ext{ V}}{35 \Omega} = 0.2 	ext{ A} = \frac{1}{5} 	ext{ A}$.$3$. The current flowing through the load resistor $R_{L} = 90 \Omega$ is $I_{L} = \frac{V_{Z}}{R_{L}} = \frac{15 	ext{ V}}{90 \Omega} = \frac{1}{6} 	ext{ A}$.$4$. The current flowing through the Zener diode is $I_{Z} = I - I_{L} = \frac{1}{5} 	ext{ A} - \frac{1}{6} 	ext{ A} = \frac{6 - 5}{30} 	ext{ A} = \frac{1}{30} 	ext{ A}$.$5$. The power dissipated across the Zener diode is $P = V_{Z} 	imes I_{Z} = 15 	ext{ V} 	imes \frac{1}{30} 	ext{ A} = 0.5 	ext{ W}$.$6$. Expressing this in the form $x 	imes 10^{-1} 	ext{ W}$, we get $0.5 	ext{ W} = 5 	imes 10^{-1} 	ext{ W}$.$7$. Therefore, the value of $x$ is $5$.

The value of power dissipated across the Zener diode $(V_{Z} = 15 \text{ V})$ connected in the circuit as shown in the figure is $x \times 10^{-1} \text{ watt}$. The value of $x$, to the nearest integer, is ..............

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