In a given circuit diagram- a 5 V zener diode along with a series resistance is connected across a 50 V power supply- The minimum value of the resistance required- if the maximum Zener current is 90 mA will be - Integer only

Voltage across, nbsp; R_v is nbsp; 5 V. ⇒ I_2 =5R_v Also voltage across nbsp; R, V_R nbsp; = 50 – 5 = 45 V Using, V=IR⇒ R=V_RI=45I_1+I_2 R=4590× 10 ...

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Physics

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In a given ci...

Question

In a given circuit diagram, a $5~\text V$ zener diode along with a series resistance is connected across a $50~\text V$ power supply. The minimum value of the resistance required, if the maximum Zener current is $90 ~\text{mA} $ will be $\Omega$. (Integer only)

V=50V

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Solution

$Voltage across, R_{v}$ is $5 V$ .

$\Rightarrow I_{2} = \frac{5}{R_{v}}$

$Also voltage across R, V_{R} = 50 - 5 = 45 V$

$Using, V = I R \Rightarrow R = \frac{V_{R}}{I} = \frac{45}{I_{1} + I_{2}}$

$R = \frac{45}{90 \times 10^{- 3} + \frac{5}{R_{v}}}$

$Current in zener diode is maximum when R_{v} \to \infty$

$(I_{2} \to 0 and I_{1} = I)$

So, $R = \frac{45}{90 \times 10^{- 3}} = 500 Ω$

Correct answer: $500$

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In a given circuit diagram, a \(5~\text V\) zener diode along with a series resistance is connected across a \(50~\text V\) power supply. The minimum value of the resistance required, if the maximum Zener current is \(90 ~\text{mA} \) will be \(\Omega\). (Integer only) V=50V

Voltage across,Rv is 5 V. ⇒I2=5Rv Also voltage across R,VR=50–5=45 V Using, V=IR⇒R=VRI=45I1+I2 R=4590×10−3+5Rv Current in zener diode is maximum when Rv→∞ (I2→0 and I1=I) So, R=4590×10−3=500 Ω Correct answer: 500

In a given circuit diagram, a \(5~\text V\) zener diode along with a series resistance is connected across a \(50~\text V\) power supply. The minimum value of the resistance required, if the maximum Zener current is \(90 ~\text{mA} \) will be \(\Omega\). (Integer only)

V=50V