The cut-off voltage of the diodes (shown in f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In the given circuit,diode $D_1$ is forward-biased,while diode $D_2$ is reverse-biased. Therefore,no current flows through the branch containing $

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) In the given circuit,diode $D_1$ is forward-biased,while diode $D_2$ is reverse-biased. Therefore,no current flows through the branch containing $D_2$.The circuit simplifies to a series combination of the $1 \,V$ battery,the $60 \,\Omega$ resistor,the forward-biased diode $D_1$ (with a voltage drop of $0.6 \,V$),and the $40 \,\Omega$ resistor.Applying Kirchhoff's Voltage Law $(KVL)$ in the loop:$1 - I(60) - 0.6 - I(40) = 0$Simplifying the equation:$0.4 - I(100) = 0$$I(100) = 0.4$$I = \frac{0.4}{100} \,A$$I = 0.004 \,A = 4 \,mA$Thus,the current through the $40 \,\Omega$ resistor is $4 \,mA$.

The cut-off voltage of the diodes (shown in figure) in forward bias is $0.6 \,V$. The current through the resistor of $40 \,\Omega$ is ........... $mA$.

Similar Questions

When a forward bias is applied to a $p-n$ junction,it

If each diode has a forward bias resistance of $25\,\Omega$ in the circuit shown below,which of the following options is correct?

The diode voltage is $0.5 \ V$ and the maximum power rating is $100 \ mW$. To obtain the maximum current in the circuit,the value of $R$ should be .......... $\Omega$.

$A$ circuit is arranged as shown in the figure. The output voltage ${V}_{0}$ is equal to $\ldots \ldots \text{ V}$.

Two ideal diodes are connected in the network as shown in the figure. The equivalent resistance between $A$ and $B$ is $.......\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module