A circular coil has a resistance of `40 Omega`. Two points P and Q of the coil, which are one quarter of the circumference apart are connected to a 16 V battery, having an internal resistance of `0.5 Omega` ? What is the main current flowing in the circuit ?

Options

A. 0.5 A
B. 1A
C. 2A
D. 3A

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the resistances of the segments of the coil
The total resistance of the coil is given as \( R = 40 \, \Omega \). Since points P and Q are one quarter of the circumference apart, we can divide the total resistance into two parts:
- \( R_1 \) (the resistance between points P and Q) will be \( \frac{R}{4} = \frac{40 \, \Omega}{4} = 10 \, \Omega \)
- \( R_2 \) (the remaining part of the coil) will be \( R - R_1 = 40 \, \Omega - 10 \, \Omega = 30 \, \Omega \)

### Step 2: Calculate the equivalent resistance of the circuit
The resistances \( R_1 \) and \( R_2 \) are in parallel with the internal resistance of the battery \( r = 0.5 \, \Omega \). The formula for the equivalent resistance \( R_{eq} \) of two resistors in parallel is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{30} \]
Finding a common denominator (which is 30):
\[ \frac{1}{R_{eq}} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} = \frac{2}{15} \]
Thus,
\[ R_{eq} = \frac{15}{2} = 7.5 \, \Omega \]

### Step 3: Find the total resistance in the circuit
Now, we need to add the internal resistance of the battery to the equivalent resistance:
\[ R_{total} = R_{eq} + r = 7.5 \, \Omega + 0.5 \, \Omega = 8 \, \Omega \]

### Step 4: Calculate the current using Ohm's Law
Using Ohm's Law, \( V = I \cdot R \), we can find the current \( I \):
\[ I = \frac{V}{R_{total}} = \frac{16 \, V}{8 \, \Omega} = 2 \, A \]

### Final Answer
The main current flowing in the circuit is \( 2 \, A \).

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