Modern Physics 6 Question 11

11. The circuit shown below contains two ideal diodes, each with a forward resistance of 50Ω. If the battery voltage is 6V, the current through the 100Ω resistance (in ampere) is

(Main 2019, 11 Jan II)

(a) 0.027

(b) 0.020

(c) 0.030

(d) 0.036

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Solution:

  1. In this circuit, D1 is forward biased and D2 is reversed biased.

Resistance of D1 is 50Ω.

Net resistance of the circuit,

Rnet =50+150+100=300Ω

Current through the 100Ω resistance

=VRnet =6300=0.020A

Key Idea When the applied reverse voltage (V) reaches the breakdown voltage of the Zener diode, then only a large amount of current is flown through it, otherwise it is approximately zero.

In the given situation, if we consider that Zener diode is at breakdown. Then, potential drop across 1500Ω resistances will be 10V. So potential drop at 500Ω resistor will be 2V.

Current in R1=2500=4mA=I1 (say)

Current in each

R2=10750=2150=13.33mA=I2 (say) I1<I2 which is not possible. 

So, Zener diode will never reach to its breakdown.

Current flowing through a reverse biased Zener diode =0



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