JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Current Electricity

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of

40Ω

is ______ mA.

Answer: 4

Solution

Applying Kirchhoff's voltage law around the circuit:

1 - I (60) - 0.6 - I (40) = 0

The diode drops 0.6 V when forward biased. Simplifying:

1 - 0.6 = I (60 + 40)

0.4 = 100 I

I = \frac{0.4}{100} = 0.004 A = 4 mA

The current through the

40Ω

resistor is 4 mA.

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer