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If current in diode is five times that in $R_1$ and breakdown voltage of diode is $6 \mathrm{~V}$, find the value of $R$ ?
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Verified Answer
The correct answer is:
$\frac{2000}{3} \Omega$
Given that, $R_{\mathrm{l}}=1 \mathrm{k} \Omega$,
Zener breakdown voltage $V_z=6 \mathrm{~V}$
Let $I_1$ and $I_z$ be the current passing through $R_1$ and Zener diode, then using equation,
$I_1=\frac{V_z}{R_1}=\frac{6}{1000} \mathrm{~A}$
According to given condition Zener current,
$I_z=5 I_1=\frac{30}{1000} \mathrm{~A}$
Let $I$ be the current drawn from source battery $\left(V_s=30 \mathrm{~V}\right)$, then
$I=\frac{V_{\mathrm{s}}-V_z}{R}=I_1+I_z$
$\frac{30-6}{R}=\frac{6}{1000}+\frac{30}{1000}$
$\Rightarrow \quad R=\frac{2000}{3} \Omega$
Zener breakdown voltage $V_z=6 \mathrm{~V}$
Let $I_1$ and $I_z$ be the current passing through $R_1$ and Zener diode, then using equation,
$I_1=\frac{V_z}{R_1}=\frac{6}{1000} \mathrm{~A}$
According to given condition Zener current,
$I_z=5 I_1=\frac{30}{1000} \mathrm{~A}$
Let $I$ be the current drawn from source battery $\left(V_s=30 \mathrm{~V}\right)$, then
$I=\frac{V_{\mathrm{s}}-V_z}{R}=I_1+I_z$
$\frac{30-6}{R}=\frac{6}{1000}+\frac{30}{1000}$
$\Rightarrow \quad R=\frac{2000}{3} \Omega$
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