The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit ?

Question

The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit ? (A) 1.71 A (B) 2.00 A (C) 2.31 A (D) 1.33 A

Tardigrade Admin · Accepted Answer

p-n Junction Diode offers negligible resistance in forward biasing and offers very large resistance in reverse biasing. Step1: Identify the Diode in reverse biasing and remove that branch.
The diode

D_{1}

is in reverse biasing thus that branch can be neglected. The diode

D_{2}

is in forward biasing and it offers negligible resistance.
Step 2: Calculate net resistance of circuit and the find current. Equivalent resistance of the circuit is,

R_{eq.} = (4 + 2) Ω \Rightarrow R_{eq.} = 6Ω

Thus current flowing in the circuit is, by Ohm's Law

I = \frac{V}{R _{eq}} \Rightarrow I = \frac{12}{6} \Rightarrow I = 2 A

Thus, net current flowing through the circuit is

2 A

.

Q. The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit ?

$1.71 A$

$2.00 A$

$2.31 A$

$1.33 A$

Q. The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit ?

1.71A

2.00A

2.31A

1.33A

$1.71 A$

$2.00 A$

$2.31 A$

$1.33 A$