If the current in the Zener diode is five tim | vedclass

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(B) Given that,$R_1 = 1 	ext{ k}\Omega = 1000 	ext{ } \Omega$,and Zener breakdown voltage $V_z = 6 	ext{ V}$.Let $I_1$ be the current through $R_1$

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$\frac{2000}{3} 	ext{ } \Omega$

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(B) Given that,$R_1 = 1 	ext{ k}\Omega = 1000 	ext{ } \Omega$,and Zener breakdown voltage $V_z = 6 	ext{ V}$.Let $I_1$ be the current through $R_1$ and $I_z$ be the current through the Zener diode.Since $R_1$ is in parallel with the Zener diode,the voltage across $R_1$ is $V_z = 6 	ext{ V}$.Therefore,$I_1 = \frac{V_z}{R_1} = \frac{6 	ext{ V}}{1000 	ext{ } \Omega} = 6 	imes 10^{-3} 	ext{ A}$.According to the given condition,the Zener current is $I_z = 5 I_1 = 5 	imes (6 	imes 10^{-3} 	ext{ A}) = 30 	imes 10^{-3} 	ext{ A}$.The total current $I$ drawn from the source $V_s = 30 	ext{ V}$ is $I = I_1 + I_z = 6 	imes 10^{-3} + 30 	imes 10^{-3} = 36 	imes 10^{-3} 	ext{ A}$.Also,the voltage drop across $R$ is $V_s - V_z = 30 	ext{ V} - 6 	ext{ V} = 24 	ext{ V}$.Using Ohm's law,$R = \frac{V_s - V_z}{I} = \frac{24}{36 	imes 10^{-3}} = \frac{24000}{36} 	ext{ } \Omega = \frac{2000}{3} 	ext{ } \Omega$.

If the current in the Zener diode is five times the current in $R_1$ and the breakdown voltage of the diode is $6 \text{ V}$,find the value of $R$.

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