A power supply has internal resistance RS and open load voltage VS = 5 V. When
a load resistance R is connected to the power supply, a voltage drop of V = 4 V
is measured across the load. The value of
is __________ (Round off to the
nearest integer)
Official Answer Key:
4 to 4
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.30)**
**1. Circuit Principles & Voltage Divider**:
- Open-circuit source voltage: $V_S = 5$ V.
- Internal resistance: $R_S$.
- Load resistance: $R_L$.
- Measured load voltage: $V_L = 4$ V.
**2. Detailed Step-by-Step Derivation**:
- The circuit forms a simple series loop where current $I$ is:
$I = \frac{V_S}{R_S + R_L}$
- The voltage drop across the load is:
$V_L = I R_L = \frac{V_S R_L}{R_S + R_L}$
- Substitute the known values:
$4 = \frac{5 R_L}{R_S + R_L}$
$4(R_S + R_L) = 5 R_L$
$4R_S + 4R_L = 5R_L$
$4R_S = R_L$
- Taking the ratio:
$\frac{R_L}{R_S} = 4$
**3. Official Verification**:
The verified official IIT Master Answer range is **4 to 4**.
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