KCET · Physics · Semiconductors
A \(p-n\) junction diode is connected to a battery of emf 5.7 V in series with a resistant \(5 \mathrm{k} \Omega\) such that it is forward biased. If the barrier potential of the diode is 0.7 V , neglecting the diode resistance, the current in the circuit is
- A 1.14 mA
- B 1 mA
- C 1 A
- D 1.14 A
Answer & Solution
Correct Answer
(B) 1 mA
Step-by-step Solution
Detailed explanation
Given, barrier potential, \(V_B=0.7 \mathrm{~V}\)
Series resistance,
\(R_S=5 \mathrm{k} \Omega=5 \times 10^3 \Omega\)
\(\therefore \quad I=\frac{V-V_B}{R_S}=\frac{5.7-0.7}{5 \times 10^3}=1 \times 10^{-3} \mathrm{~A}=1 \mathrm{~mA}\)
Series resistance,
\(R_S=5 \mathrm{k} \Omega=5 \times 10^3 \Omega\)
\(\therefore \quad I=\frac{V-V_B}{R_S}=\frac{5.7-0.7}{5 \times 10^3}=1 \times 10^{-3} \mathrm{~A}=1 \mathrm{~mA}\)
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