AP EAMCET · PHYSICS · Semiconductors
A Zener diode voltage regulator operated in the range \(120 \mathrm{~V}-180 \mathrm{~V}\) produces a constant supply of \(110 \mathrm{~V}\) and \(250 \mathrm{~mA}\) to the load. If the maximum current is equally shared between the load and the Zener diode, then the values of load resistance \(\left(R_L\right)\) and series resistance \(\left(R_S\right)\) are respectively
- A \(R_L=280 \Omega, R_S=70 \Omega\)
- B \(R_L=440 \Omega, R_S=140 \Omega\)
- C \(R_L=70 \Omega, R_S=280 \Omega\)
- D \(R_L=440 \Omega, R_S=1400 \Omega\)
Answer & Solution
Correct Answer
(B) \(R_L=440 \Omega, R_S=140 \Omega\)
Step-by-step Solution
Detailed explanation
We have, \(R_L=\frac{V_L}{I_L}=\frac{110}{250 \times 10^{-3}}=440 \Omega\) As current is equally shared by Zener diode and resistance \(R\). So, \(I_Z=250 \mathrm{~mA}\) Then, \(I=2 \times 250=500 \mathrm{~mA} \quad\left[\because I=I_L+I_Z\right]\) \(=0.5 \mathrm{~mA}\) So,…
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