The value of power dissipated across the zener (diode Vz=15V)

Question:

The value of power dissipated across the zener diode $\left(\mathrm{V}_{\mathrm{z}}=15 \mathrm{~V}\right)$ connected in the circuit as shown in the figure is $x \times 10^{-1}$ watt.

The value of $x$, to the nearest integer, is

Solution:

(5)

Voltage across $\mathrm{R}_{\mathrm{S}}=22-15=7 \mathrm{~V}$

Current through $\mathrm{R}_{\mathrm{S}}=\mathrm{I}=\frac{7}{35}=\frac{1}{5} \mathrm{~A}$

Current through $90 \Omega=\mathrm{I}_{2}=\frac{15}{90}=\frac{1}{6} \mathrm{~A}$

Current through zener $=\frac{1}{5}-\frac{1}{6}=\frac{1}{30} \mathrm{~A}$

Power through zener diode

$\mathrm{P}=\mathrm{VI}$

$\mathrm{P}=15 \times \frac{1}{30}=0.5 \mathrm{watt}$

$\mathrm{P}=5 \times 10^{-1}$ watt

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