KCET · Physics · Nuclear Physics
The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in
the circuit?
- A \( 1.71 \mathrm{~A} \)
- B ( \( 2.0 \mathrm{~A} \)
- C \( 2.31 \mathrm{~A} \)
- D \( 1.33 \mathrm{~A} \)
Answer & Solution
Correct Answer
(A) \( 1.71 \mathrm{~A} \)
Step-by-step Solution
Detailed explanation
Here, diode \( D_{1} \) is forward biased and \( D_{2} \) is reverse biased, therefore, diode \( D_{2} \) will not conduct any electricity.
Current
\[
I=\frac{V}{R_{2}+R_{3}}=\frac{12}{3 \Omega+4 \Omega}=\frac{12}{7}=1.71 A
\]
Current
\[
I=\frac{V}{R_{2}+R_{3}}=\frac{12}{3 \Omega+4 \Omega}=\frac{12}{7}=1.71 A
\]
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