A cell of emf. `4.5 V` is connected to a junction diode whose barrier potential is `0.7 V`. If the external resistance in the circuit is `190 Omega`. The current in the circuit is

Options

A. `20 mA`
B. `2 mA`
C. `23 mA`
D. `200 mA`

Explanation

To find the current in the circuit with the given parameters, we can follow these steps:

### Step 1: Understand the Circuit Configuration
We have a battery with an EMF of 4.5 V connected in series with a diode (which has a barrier potential of 0.7 V) and an external resistor of 190 Ω. The diode is forward-biased.

### Step 2: Apply Kirchhoff’s Voltage Law (KVL)
In a closed loop, the sum of the potential differences (voltage) must equal zero. The equation can be set up as follows:

\[ \text{EMF} - \text{Voltage drop across diode} - \text{Voltage drop across resistor} = 0 \]

Substituting the known values:

\[ 4.5 V - 0.7 V - (I \times 190 \, \Omega) = 0 \]

### Step 3: Simplify the Equation
Rearranging the equation gives:

\[ 4.5 V - 0.7 V = I \times 190 \, \Omega \]

Calculating the left side:

\[ 3.8 V = I \times 190 \, \Omega \]

### Step 4: Solve for Current (I)
Now, we can isolate I:

\[ I = \frac{3.8 V}{190 \, \Omega} \]

### Step 5: Perform the Calculation
Calculating the current:

\[ I = \frac{3.8}{190} = 0.02 \, \text{A} = 20 \, \text{mA} \]

### Final Answer
The current in the circuit is **20 mA**.

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