A Zener diode with a breakdown voltage of 4 V is connceted in sereis with a resistance R to a battery of emf 10 V. The maximum power dissipation rating for the Zener diode is 1 W. The value of R to ensure maximum power dissipation across the diode is

Options

A. `12Omega`
B. `24Omega`
C. `36Omega`
D. `6Omega`

Answer

To find the value of resistance \( R \) that ensures maximum power dissipation across the Zener diode, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Breakdown voltage of Zener diode, \( V_Z = 4 \, \text{V} \)
- EMF of the battery, \( E = 10 \, \text{V} \)
- Maximum power dissipation of the Zener diode, \( P_Z = 1 \, \text{W} \)

2. **Calculate the Current through the Zener Diode:**
The power dissipated by the Zener diode can be expressed using the formula:
\[ P_Z = V_Z \cdot I_Z \]
Rearranging this gives:
\[ I_Z = \frac{P_Z}{V_Z} \]
Substituting the known values:
\[ I_Z = \frac{1 \, \text{W}}{4 \, \text{V}} = \frac{1}{4} \, \text{A} \]

3. **Apply Kirchhoff's Voltage Law:**
According to Kirchhoff's law, the sum of the voltage drops in a closed loop must equal the EMF. The voltage drop across the resistor \( R \) can be expressed as:
\[ E - V_Z = I_Z \cdot R \]
Substituting the known values:
\[ 10 \, \text{V} - 4 \, \text{V} = \left(\frac{1}{4} \, \text{A}\right) \cdot R \]
Simplifying this gives:
\[ 6 \, \text{V} = \left(\frac{1}{4} \, \text{A}\right) \cdot R \]

4. **Solve for Resistance \( R \):**
Rearranging the equation to solve for \( R \):
\[ R = \frac{6 \, \text{V}}{\frac{1}{4} \, \text{A}} = 6 \, \text{V} \cdot 4 \, \text{A}^{-1} = 24 \, \Omega \]

### Final Answer:
The value of resistance \( R \) to ensure maximum power dissipation across the diode is \( R = 24 \, \Omega \).

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