The reverse bias in a junction diode is changed from $5 V$ to $15 V$ then the value of current changes from $38$ $μ A$ to $88$ $μ A$ . The resistance of junction diode will be
$4 \times 10^{5} Ω$
$2 \times 10^{5} Ω$
$10^{6} Ω$
$3 \times 10^{5} Ω$

Question

The reverse bias in a junction diode is changed from $5 V$ to $15 V$ then the value of current changes from $38$ $μ A$ to $88$ $μ A$ . The resistance of junction diode will be
$4 \times 10^{5} Ω$
$2 \times 10^{5} Ω$
$10^{6} Ω$
$3 \times 10^{5} Ω$

Toppr Admin · Accepted Answer

Given that-Change in voltage- V-15-x2212-5-10-xA0-VChange in current- I-88-x2212-38-50-x3BC-AHence- resistance of the diode isR-VI-1050-xD7-10-x2212-6R-0-2-xD7-106-2-xD7-105

The reverse bias in a junction diode is changed from 5 V to 15 V then the value of current changes from 38μA to 88μA. The resistance of junction diode will be4×105Ω2×105Ω106Ω3×105Ω

Given that:Change in voltage, V=15−5=10 VChange in current, I=88−38=50μAHence, resistance of the diode isR=VI=1050×10−6R=0.2×106=2×105

The reverse bias in a junction diode is changed from $5 V$ to $15 V$ then the value of current changes from $38$ $μ A$ to $88$ $μ A$ . The resistance of junction diode will be
$4 \times 10^{5} Ω$
$2 \times 10^{5} Ω$
$10^{6} Ω$
$3 \times 10^{5} Ω$

Given that:
Change in voltage, $V = 15 - 5 = 10 V$
Change in current, $I = 88 - 38 = 50$ $μ A$
Hence, resistance of the diode is
$R = \frac{V}{I} = \frac{10}{50 \times 10^{- 6}}$

$R = 0.2 \times 10^{6} = 2 \times 10^{5}$