The diode shown in the circuit is a silicon d | vedclass

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

Question

(A) In the given circuit,the positive terminal of the $6 \ V$ battery is connected to the $n$-side of the diode (cathode),and the negative terminal is

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) In the given circuit,the positive terminal of the $6 \ V$ battery is connected to the $n$-side of the diode (cathode),and the negative terminal is connected to the $p$-side through the resistor. This configuration means the diode is in reverse bias. In reverse bias,an ideal diode acts as an open circuit,and a real diode acts as a very high resistance. Since there is no current flowing through the circuit,there is no voltage drop across the $2 \ \Omega$ resistor. Therefore,the potential at point $A$ is equal to the potential of the positive terminal of the battery,which is $6 \ V$. Point $B$ is connected to the ground,so its potential is $0 \ V$. The potential difference between points $A$ and $B$ is $V_A - V_B = 6 \ V - 0 \ V = 6 \ V$.

The diode shown in the circuit is a silicon diode. The potential difference between the points $A$ and $B$ will be.....$V$

Similar Questions

The cause of the potential barrier in a $p-n$ junction diode is

$A$ diode is connected in parallel with a resistance $R$ as shown in the figure. The most probable current $(I)$ - voltage $(V)$ characteristic is:

In the diagram shown, the resistance between points $A$ and $B$ is '$R_1$' when an ideal diode $D$ is forward biased and is '$R_2$' when ideal diode $D$ is reverse biased. The ratio $R_1/R_2$ is

The $V-I$ characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is

For an ideal diode,the current in the following arrangement is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module