KCET · Physics · Semiconductors
In the diagram shown, the Zener diode has a reverse breakdown voltage of \(V_Z\). The current through the load resistance \(R_L\) is \(I_L\). The current through the Zener diode is
- A \(\frac{V_0-V_Z}{R_S}\)
- B \(\frac{V_0-V_Z}{R_L}\)
- C \(\frac{V_Z}{R_L}\)
- D \(\left(\frac{V_0-V_Z}{R_S}\right)-I_L\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{V_0-V_Z}{R_S}\right)-I_L\)
Step-by-step Solution
Detailed explanation
From the diagram,
\(I=I_Z+I_L\)
\(\Rightarrow \quad I_Z=I-I_L=\frac{V_O-V_Z}{R_S}-I_L\)
\(I=I_Z+I_L\)
\(\Rightarrow \quad I_Z=I-I_L=\frac{V_O-V_Z}{R_S}-I_L\)
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