In the circuit shown below, the maximum power dissipated by R_{s} if input varies from 15 V to 20 V (assuming Zener diode used is 10 V, 2W) is?

This question was previously asked in

TANGEDCO AE EC 2015 Official Paper

Option 2 : 2200 mW

CT 1: Determinants and Matrices

2903

10 Questions
10 Marks
12 Mins

In the given question, Supply is varying. Hence the load current is constant.

The maximum Zener current (I_{Z(max)}) is defined as:

\(I_{Z(max)}=\frac{power~rating}{Zener ~voltage}\)

Given power rating = 2 W.

\(I_{Z(max)}=\frac{2}{10}=0.2~A\)

The current across the load I_{L} will be constant to the value:

\(I_L=\frac{V_Z}{R_L}=\frac{10}{500}\)

I_{L} = 0.02 A

Now, the maximum current across the resistor R_{s} will be:

I_{S(max)} = I_{L }+ I_{Z(max)}

IS(max) = 0.02 + 0.2 = 0.22 A

The maximum voltage across R_{s} will be:

V_{S(max)} = V_{i(max)} - V_{Z} = 20 - 10 = 10 V

VS(max) = 10 V

∴ The maximum power that can be dissipated across the R_{s} will be:

P = V_{max} I_{s}(max) = (20 - 10) × 0.22 W

= 2.2 W = 2200 mW