JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
The circuit contains two diodes each with a forward resistance of \(50\, \Omega\) and with infinite reverse resistance. If the battery voltage is \(6\, V\), the current through the \(120\, \Omega\) resistance is \(mA\)
- A \(20\)
- B \(40\)
- C \(10\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(20\)
Step-by-step Solution
Detailed explanation
In this circuit \(D _{1}\) will be forward bias and \(D _{2}\) will be revers bias. \(\therefore\) There will be no current through \(D _{2}\) and \(R _{2}\) Apply \(KVL\) in circuit we get \(+6-50 i -130 i -120 i =0\) \(i =\frac{6}{300} A =\frac{6}{300} \times 1000 mA\)…
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