Currents flowing in each of the following cir | vedclass

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

Question

(C) In circuit $A$,both $p-n$ junction diodes are in forward bias. Therefore,current flows through both branches.The equivalent resistance $R_A$ is gi

Vedclass · Accepted Answer

$4 	ext{ A}, 2 	ext{ A}$

Vedclass · Answer

(C) In circuit $A$,both $p-n$ junction diodes are in forward bias. Therefore,current flows through both branches.The equivalent resistance $R_A$ is given by:$\frac{1}{R_A} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies R_A = 2 \Omega$Using Ohm's law,$V = I_A R_A$:$8 = I_A 	imes 2 \implies I_A = 4 	ext{ A}$In circuit $B$,the upper diode is in forward bias,but the lower diode is in reverse bias. Therefore,no current flows through the lower branch.The resistance of the circuit is $R_B = 4 \Omega$.Using Ohm's law,$V = I_B R_B$:$8 = I_B 	imes 4 \implies I_B = 2 	ext{ A}$Thus,the currents are $4 	ext{ A}$ and $2 	ext{ A}$ respectively.

Currents flowing in each of the following circuits $A$ and $B$ respectively are:

Similar Questions

If $A$ is a square matrix such that $A(\operatorname{adj} A) = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,then $\operatorname{det}(\operatorname{adj} A)$ is equal to

If the curves $x^2+p y^2=1$ and $q x^2+y^2=1$ are orthogonal to each other, then

If $\alpha, \beta$ are the roots of $x^2 - px + q = 0$ and $\alpha', \beta'$ are the roots of $x^2 - p'x + q' = 0$,then find the value of $(\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$.

$A$ distinctive and characteristic functional group in fats is

Choose the wrong statement:

Vedclass Test Series

Exam Paper Generator

Online Exam Module