Question

A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).

Hint:

To solve such problems, use Ohm’s law to relate voltage, current, and resistance, and apply the concept of internal resistance of the battery.

Answer 1

Step 1: Establish the equation for the scenario with current = 2A

The potential difference across the rheostat is provided as:

V₁ = E - (2 × r) = 5V

This yields the equation:

E - 2r = 5

Step 2: Establish the equation for the scenario with current = 4A

The potential difference across the rheostat in this case is:

V₂ = E - (4 × r) = 4V

This yields the equation:

E - 4r = 4

Step 3: Resolve the system of equations

The resulting system of equations is:

  E - 2r = 5
  E - 4r = 4
  

Subtracting the second equation from the first yields:

2r = 1

Consequently, r = 1/2 Ω.

Step 4: Substitute the determined value of r into one of the equations

Substituting r = 1/2 Ω into the first equation:

E - 1 = 5

Therefore, E = 6V.

Final Answer:

EMF of the battery: E = 6V

Internal resistance of the battery: r = 1/2 Ω