In the following circuit,the current flowing | vedclass

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(B) The Zener diode is connected in parallel with resistor $R_2$. Therefore,the voltage across $R_2$ is equal to the Zener breakdown voltage,$V_Z = 8

Vedclass · Accepted Answer

$25$

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(B) The Zener diode is connected in parallel with resistor $R_2$. Therefore,the voltage across $R_2$ is equal to the Zener breakdown voltage,$V_Z = 8 	ext{ V}$.The current flowing through resistor $R_2$ is:$I_{R_2} = \frac{V_Z}{R_2} = \frac{8 	ext{ V}}{1600 	ext{ }\Omega} = 5 	imes 10^{-3} 	ext{ A} = 5 	ext{ mA} \quad \dots(i)$The total voltage supplied is $20 	ext{ V}$. The voltage drop across resistor $R_1$ is:$V_{R_1} = V_{	ext{source}} - V_Z = 20 	ext{ V} - 8 	ext{ V} = 12 	ext{ V}$The total current flowing through the circuit (which passes through $R_1$) is:$I_{R_1} = \frac{V_{R_1}}{R_1} = \frac{12 	ext{ V}}{400 	ext{ }\Omega} = 3 	imes 10^{-2} 	ext{ A} = 30 	ext{ mA} \quad \dots(ii)$Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $I_Z$ is:$I_Z = I_{R_1} - I_{R_2}$$I_Z = 30 	ext{ mA} - 5 	ext{ mA} = 25 	ext{ mA}$

In the following circuit,the current flowing through the Zener diode is (in $\text{ mA}$)

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