Two resistors of resistances `2Omega` and `6Omega` are connected in parallel. This combination is then connected to a battery of emf 2 V and internal resistance `0.5 Omega`. What is the current flowing through the battery ?

Options

A. `4 A`
B. `(4)/(3)A`
C. `(4)/(17)A`
D. `1 A`

Answer

To solve the problem, we need to find the current flowing through the battery when two resistors of 2Ω and 6Ω are connected in parallel, and this combination is connected to a battery with an EMF of 2V and an internal resistance of 0.5Ω.

### Step 1: Calculate the equivalent resistance of the parallel resistors

The formula for the equivalent resistance \( R_p \) of two resistors \( R_1 \) and \( R_2 \) connected in parallel is given by:

\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \]

Substituting the values \( R_1 = 2Ω \) and \( R_2 = 6Ω \):

\[ \frac{1}{R_p} = \frac{1}{2} + \frac{1}{6} \]

Finding a common denominator (which is 6):

\[ \frac{1}{R_p} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \]

Now, taking the reciprocal to find \( R_p \):

\[ R_p = \frac{3}{2}Ω = 1.5Ω \]

### Step 2: Calculate the total equivalent resistance in the circuit

The total equivalent resistance \( R \) in the circuit is the sum of the internal resistance \( R_i \) and the equivalent resistance of the parallel combination \( R_p \):

\[ R = R_i + R_p \]

Substituting the values \( R_i = 0.5Ω \) and \( R_p = 1.5Ω \):

\[ R = 0.5 + 1.5 = 2Ω \]

### Step 3: Calculate the current flowing through the battery

Using Ohm's law, the current \( I \) flowing through the circuit can be calculated using the formula:

\[ I = \frac{E}{R} \]

Where \( E \) is the EMF of the battery. Substituting \( E = 2V \) and \( R = 2Ω \):

\[ I = \frac{2}{2} = 1A \]

### Final Answer

The current flowing through the battery is **1 ampere**.

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