In an ideal junction diode, the current flowi | vedclass

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(C) The diode is connected in forward bias because the $p$-side is at a higher potential $(+3 	ext{ V})$ than the $n$-side $(-5 	ext{ V})$.For an id

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$4 	imes 10^{-3} 	ext{ A}$

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(C) The diode is connected in forward bias because the $p$-side is at a higher potential $(+3 	ext{ V})$ than the $n$-side $(-5 	ext{ V})$.For an ideal junction diode in forward bias, the resistance of the diode is zero.The potential difference across the resistor is $V = V_P - V_Q = 3 	ext{ V} - (-5 	ext{ V}) = 8 	ext{ V}$.The resistance $R = 2 	ext{ k}\Omega = 2000 	ext{ }\Omega$.Using Ohm's law, the current $I$ is given by $I = \frac{V}{R} = \frac{8 	ext{ V}}{2000 	ext{ }\Omega} = 4 	imes 10^{-3} 	ext{ A}$.

In an ideal junction diode, the current flowing through $PQ$ is (resistance is $2 \text{ k}\Omega$).

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