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In the circuit shown in Fig. 14.5, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is


A. 1.3 V
B. 2.3 V
C. 0
D. 0.5 V

Answers (1)

 The answer is an option (b)

Let us consider fig (b)given above in the problem suppose the potential difference between A and B is VAB

Then,

\\V_{AB}-0.3= [ ( r_{1} +r_{2}) 10^{3} ]\times ( 0.2 \times 10^{-3} ) [ \ because\ V_{AB}=ir ]\\ = [( 5+5 ) 10^{3}]\times ( 0.2 \times 10^{-3} )\\ =10\times 10^{3}\times 0.2\times 10^{-3}-2\\ \Rightarrow V_{AB}=2+0.3=2.3V

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