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) 2.0 A (C) 2.31 A (D) 1.33 A

Tardigrade Admin · Accepted Answer

In the given circuit, D1 is forward biased and D2 is reverse biased. So, D1 works like a closed switch and D2 works like an open circuit. So, circuit becomes

Now, current in the circuit, I=(E/Rs)=(12/4+3) I=(12/7) I=1.71 A

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

$1.71 A$

$2.0 A$

$2.31 A$

$1.33 A$

In the given circuit, $D_{1}$ is forward biased and $D_{2}$ is reverse biased. So, $D_{1}$ works like a closed switch and $D_{2}$ works like an open circuit. So, circuit becomes

Now, current in the circuit, $I = \frac{E}{R _{s}} = \frac{12}{4 + 3}$
$I = \frac{12}{7}$
$I = 1.71 A$

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

1.71A

2.0A

2.31A

1.33A

In the given circuit, D1​ is forward biased and D2​ is reverse biased. So, D1​ works like a closed switch and D2​ works like an open circuit. So, circuit becomes Now, current in the circuit, I=Rs​E​=4+312​ I=712​ I=1.71A

$1.71 A$

$2.0 A$

$2.31 A$

$1.33 A$

In the given circuit, $D_{1}$ is forward biased and $D_{2}$ is reverse biased. So, $D_{1}$ works like a closed switch and $D_{2}$ works like an open circuit. So, circuit becomes

Now, current in the circuit, $I = \frac{E}{R _{s}} = \frac{12}{4 + 3}$
$I = \frac{12}{7}$
$I = 1.71 A$