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

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(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

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$6$

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(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.....$V$

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