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

Options

A. 20mA
B. 2mA
C. 0.2mA
D. 0.02mA

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the circuit components
We have a cell with an electromotive force (emf) of 4.5V, a junction diode with a barrier potential of 0.7V, and an external resistance of 190Ω.

### Step 2: Write down the formula for the circuit
According to Kirchhoff's Voltage Law (KVL), the sum of the voltages around a closed loop must equal zero. The equation can be set up as follows:

\[ \text{emf} - \text{barrier potential} - I \cdot R = 0 \]

Where:
- emf = 4.5V
- barrier potential = 0.7V
- I = current in the circuit
- R = resistance = 190Ω

### Step 3: Substitute the known values into the equation
Substituting the values we have:

\[ 4.5V - 0.7V - I \cdot 190Ω = 0 \]

### Step 4: Simplify the equation
Rearranging the equation gives:

\[ 4.5V - 0.7V = I \cdot 190Ω \]

Calculating the left side:

\[ 3.8V = I \cdot 190Ω \]

### Step 5: Solve for the current (I)
Now, we can solve for I:

\[ I = \frac{3.8V}{190Ω} \]

### Step 6: Perform the calculation
Calculating the current:

\[ I = \frac{3.8}{190} \]

\[ I = 0.02A \]

### Step 7: Convert to milliamperes
To convert the current from amperes to milliamperes, we multiply by 1000:

\[ I = 0.02A \times 1000 = 20mA \]

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

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