MHT CET · Physics · Semiconductors
The given circuit has two ideal diodes \(D_1\) and \(D_2\) connected as shown in the figure. The current flowing through the resistance \(R_1\) will be
- A \(2 \mathrm{~A}\)
- B \(3.3 \mathrm{~A}\)
- C \(2.5 \mathrm{~A}\)
- D \(7 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(2.5 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
The ideal diode \(D_1\) is reversed biased whereas ideal diode \(D_2\) is forward biased. Thus, \(D_1\) acts as an open switch while \(D_2\) as a closed switch as shown in the circuit.
Thus, current flowing through \(R_1, I=\frac{V}{R_1+R_3}=\frac{10}{2+2}=2.5 \mathrm{~A}\)
Thus, current flowing through \(R_1, I=\frac{V}{R_1+R_3}=\frac{10}{2+2}=2.5 \mathrm{~A}\)
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