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(B) The diode is connected in forward bias because the $P$-terminal is at a higher potential $(+4 	ext{ V})$ than the $N$-terminal $(+1 	ext{ V})$.I

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$10^{-2} 	ext{ A}$

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(B) The diode is connected in forward bias because the $P$-terminal is at a higher potential $(+4 	ext{ V})$ than the $N$-terminal $(+1 	ext{ V})$.In forward bias,the diode acts as a closed switch (assuming an ideal diode).The potential difference across the resistor is $V = 4 	ext{ V} - 1 	ext{ V} = 3 	ext{ V}$.Using Ohm's law,the current $i$ is given by $i = \frac{V}{R} = \frac{3 	ext{ V}}{300 \ \Omega} = \frac{1}{100} 	ext{ A} = 10^{-2} 	ext{ A}$.

In the circuit given below,the value of the current is

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