The barrier potential in an ideal P-n junction diode is 0.3 volts. The current required is 6 mA. If a resistance of `200 Omega ` is connected in series with the junction diode, then the emf of the cell required for use in the circuit is

Options

A. 0.3 V
B. 1.2 V
C. 0.9 V
D. 1.5 V

Answer

To solve the problem, we need to find the electromotive force (emf) required for the circuit with a P-N junction diode, given the barrier potential, current, and resistance in the circuit.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Barrier potential (V_d) = 0.3 V
- Current (I) = 6 mA = 0.006 A (conversion from milliamperes to amperes)
- Resistance (R) = 200 Ω

2. **Use the Formula for EMF:**
The relationship between emf (E), barrier potential (V_d), current (I), and resistance (R) in a series circuit can be expressed as:
\[ E = V_d + I \cdot R \]

3. **Calculate the Voltage Drop Across the Resistor:**
First, we need to calculate the voltage drop across the resistor using Ohm's Law (V = I * R):
\[ V_R = I \cdot R = 0.006 \, \text{A} \cdot 200 \, \Omega = 1.2 \, \text{V} \]

4. **Substitute the Values into the EMF Formula:**
Now, substitute V_d and V_R into the EMF formula:
\[ E = V_d + V_R = 0.3 \, \text{V} + 1.2 \, \text{V} \]

5. **Calculate the Total EMF:**
\[ E = 0.3 + 1.2 = 1.5 \, \text{V} \]

6. **Conclusion:**
The required emf of the cell for use in the circuit is 1.5 V.

### Final Answer:
The emf required is **1.5 volts**.

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