MHT CET · Physics · Semiconductors
What are the values of the currents flowing in each of the following diode circuits \(\mathrm{X}\) and \(\mathrm{Y}\) respectively? (Assume that the diodes are ideal)
- A \(1 \mathrm{~A}, 2 \mathrm{~A}\)
- B \(2 \mathrm{~A}, 1 \mathrm{~A}\)
- C \(4 \mathrm{~A}, 2 \mathrm{~A}\)
- D \(2 \mathrm{~A}, 4 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(4 \mathrm{~A}, 2 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
In circuit \(\mathrm{X}\), both the diodes are forward biased and hence both will conduct.
The two resistances of \(4 \Omega\) each are in parallel. Their equivalent resistance is \(2 \Omega\). Hence the current \(\mathrm{I}=\frac{\mathrm{v}}{\mathrm{R}}=\frac{8}{2}=4 \mathrm{~A}\)
In the circuit \(Y\), the diode \(D_1\) is forward biased but diode \(D_2\) is reverse biased. Hence only diode \(\mathrm{D}_1\) will conduct. The resistance is \(4 \Omega\).
Hence, \(\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}\)
The two resistances of \(4 \Omega\) each are in parallel. Their equivalent resistance is \(2 \Omega\). Hence the current \(\mathrm{I}=\frac{\mathrm{v}}{\mathrm{R}}=\frac{8}{2}=4 \mathrm{~A}\)
In the circuit \(Y\), the diode \(D_1\) is forward biased but diode \(D_2\) is reverse biased. Hence only diode \(\mathrm{D}_1\) will conduct. The resistance is \(4 \Omega\).
Hence, \(\mathrm{I}=\frac{8}{4}=2 \mathrm{~A}\)
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