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Electronic Devices question

2025 · 2 Apr · Shift 1 · Q68
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Electronic Devices question

2025 · 2 Apr · Shift 1 · Q68

JEE MainPhysicsElectronic DevicesMCQ+4 / −1
A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V . For a 400Ω400 \Omega400Ω resistor connected in series, the zener current is found to be 4 times load current. The load current (IL)\left(I_L\right)(IL​) and load resistance (RL)\left(R_L\right)(RL​) are :
  1. A
    IL=0.02 mA;RL=250Ω\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \OmegaIL​=0.02 mA;RL​=250Ω
  2. B
    IL=10 A;RL=0.5Ω\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \OmegaIL​=10 A;RL​=0.5Ω
  3. C
    IL=10 mA;RL=500Ω\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \OmegaIL​=10 mA;RL​=500Ω
  4. D
    IL=20 mA;RL=250Ω\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \OmegaIL​=20 mA;RL​=250Ω
View written solutionFree

Correct answer: C

  1. Given data
  • Zener voltage: VZ=5 VV_Z = 5\,\text{V}VZ​=5V
  • Input voltage: Vin=25 VV_{in} = 25\,\text{V}Vin​=25V
  • Series resistor: Rs=400 ΩR_s = 400\,\OmegaRs​=400Ω
  • Zener current is 4 times load current: IZ=4ILI_Z = 4I_LIZ​=4IL​
  1. Current through the series resistor

Since the zener diode regulates the output, the output voltage across load is 5 V5\,\text{V}5V.

So voltage across the series resistor is VRs=Vin−VZ=25−5=20 VV_{R_s} = V_{in} - V_Z = 25 - 5 = 20\,\text{V}VRs​​=Vin​−VZ​=25−5=20V

Hence current through the series resistor is I=VRsRs=20400=0.05 A=50 mAI = \frac{V_{R_s}}{R_s} = \frac{20}{400} = 0.05\,\text{A} = 50\,\text{mA}I=Rs​VRs​​​=40020​=0.05A=50mA

  1. Apply current relation

The series current splits into zener current and load current: I=IZ+ILI = I_Z + I_LI=IZ​+IL​

Given: IZ=4ILI_Z = 4I_LIZ​=4IL​

Therefore, I=4IL+IL=5ILI = 4I_L + I_L = 5I_LI=4IL​+IL​=5IL​

So, IL=I5=50 mA5=10 mAI_L = \frac{I}{5} = \frac{50\,\text{mA}}{5} = 10\,\text{mA}IL​=5I​=550mA​=10mA

  1. Find load resistance

Load voltage is regulated at 5 V5\,\text{V}5V, so RL=VLIL=510×10−3=500 ΩR_L = \frac{V_L}{I_L} = \frac{5}{10\times10^{-3}} = 500\,\OmegaRL​=IL​VL​​=10×10−35​=500Ω

  1. Match with options

Thus, IL=10 mA,RL=500 ΩI_L = 10\,\text{mA}, \quad R_L = 500\,\OmegaIL​=10mA,RL​=500Ω

This matches Option C.

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