The above I – V graph shows the characteristics of a p-n junction diode which is connected to a signal generator of the peak voltage 22 V. The de current across the load `4.8 Omega` is į A, then i =?

Answer

A, B

Explanation

The current at 2V is 250 mA and at 2.2V it is 500 mA.
The dynamic resistance in this region is,
`r_d = (DeltaV)/(DeltaI)`
`= ((2.2 - 2))/((500 - 250) xx 10^(-3)) = 4/5 Omega`
Now, when the diode is connected to ac supply it works as a half-wave rectifier giving a dc output. For half-wave rectifier, peak current is,
`I_0 = (V_0)/(r_d + R_L)`
`= 22/(4/5 + 4.8)`
`= 22/5.6 A`
`therefore I_(dc) = (I_0)/R = (22)/(5.6 xx 22/7)`
`= 5/4 `
`= 1.25 A`