The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 Ω is mA.

Question

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 Ω is mA. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f3847b844dd63ab197bc080780a84653-.png" /> 1- I (60)-0.6- I (40)=0 (0.4/100)= I I =4 mA

Q. The cut-off voltage of the diodes (shown in figure) in forward bias is $0.6 V$ . The current through the resister of $40 Ω$ is _______ $m A$ .

$1 - I (60) - 0.6 - I (40) = 0$
$\frac{0.4}{100} = I$
$I = 4 m A$

Q. The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6V. The current through the resister of 40Ω is _______mA.

1−I(60)−0.6−I(40)=0 1000.4​=I I=4mA

Q. The cut-off voltage of the diodes (shown in figure) in forward bias is $0.6 V$ . The current through the resister of $40 Ω$ is _______ $m A$ .

$1 - I (60) - 0.6 - I (40) = 0$
$\frac{0.4}{100} = I$
$I = 4 m A$