Question

Consider the circuit given below.

Assume the diode is ideal. The average current passing through the resistor R is found to be 2mA. Then the value of resistance R is equal to________$k\Omega$

• A. 1.31

Answer 1

The correct option is A 1.31

current flows in the circuit between $t_1$ and $t_2$ only for every cycle

$24\sin \omega t_1=12$

$\omega t_1=\frac{\pi }{6}$

$\frac{2\pi }{T}{t}_1=\frac{\pi }{6}$

$t_1=\frac{T}{12}$

$t_2=\frac{T}{2}-\frac{T}{12}=\frac{5T}{12}$

$I_{avg}=\frac{1}{RT}{\int }_{t_1}^{t_2}\left[\right(24\sin \omega t)-12]dt$

$=\frac{1}{RT}\left[\frac{-24}{\omega }\right(\cos \omega t{)}_{T/12}^{5T/12}-12(t_2-t_1)]$

$=\frac{1}{RT}[\frac{24\left(\frac{\sqrt{3}}{2}\right)2}{\omega }-4T]=\frac{1}{R}[\frac{24\sqrt{3}}{2\pi }-4]$

$I_{avg}=2mA$

$R=\frac{\frac{24\sqrt{3}}{2\pi }-4}{2\times {10}^{-3}}\simeq 1.31k\Omega$