CUET Preparation Today
A junction diode is connected to a 10 V source and $10^{-3}$ resistor as shown in figure. The slope of load line on the characteristic curve of diode will be:
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$10^{-2}AV^{-1}$ $10^{-3}AV^{-1}$ $10^{-4}AV^{-1}$ $10^{-5}AV^{-1}$ |
$10^{-3}AV^{-1}$ |
If V is the voltage across the junction and I is the circuit current, then $V + IR = E$ or $I=\frac{E}{R}-\frac{V}{R}=-\frac{V}{R}+\frac{E}{R}$ Slope of load line = $-\frac{1}{R}=\frac{1}{1000}=10^{-3}AV^{-1}$ |
