In the circuit shown in figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

Question

In the circuit shown in figure, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/45e37a1c11981125f43d1d61095be701-.png" /> (A) 1.3 V (B) 2.3 V (C) 0 (D) 0.5 V

Tardigrade Admin · Accepted Answer

Let V be the potential difference between A and B, then V - 0.3 = (5 + 5) × 103× (0.2× 10-3) = 2 or V= 2 + 0.3 = 2.3 V

Q. In the circuit shown in figure, if the diode forward voltage drop is $0.3 V$ , the voltage difference between $A$ and $B$ is

$1.3 V$

$2.3 V$

$0$

$0.5 V$

Let $V$ be the potential difference between $A$ and $B$ , then
$V - 0.3 = (5 + 5) \times 1 0^{3} \times (0.2 \times 1 0^{- 3}) = 2$
or $V = 2 + 0.3 = 2.3 V$

Q. In the circuit shown in figure, if the diode forward voltage drop is 0.3V, the voltage difference between A and B is

1.3V

2.3V

0

0.5V

Let V be the potential difference between A and B, then V−0.3=(5+5)×103×(0.2×10−3)=2 or V=2+0.3=2.3V

Q. In the circuit shown in figure, if the diode forward voltage drop is $0.3 V$ , the voltage difference between $A$ and $B$ is

$1.3 V$

$2.3 V$

$0$

$0.5 V$

Let $V$ be the potential difference between $A$ and $B$ , then
$V - 0.3 = (5 + 5) \times 1 0^{3} \times (0.2 \times 1 0^{- 3}) = 2$
or $V = 2 + 0.3 = 2.3 V$