Question

A battery of 10 V and internal resistance 0.5 Ω is connected in parallel with a battery of 12 V and internal resistance 0.8 Ω. The terminals are connected by an external resistance of 20 Ω. The current flowing through the 20 Ω resistance is:

• A. 0.75 A
• B. 1.74 A
• C. 0.53 A
• D. 1.21 A

Hint:

When multiple batteries are connected in parallel, their equivalent EMF and internal resistance must be calculated before determining the total current in the circuit.

Answer 1

The Correct Option is C

To find the current flowing through the 20 Ω resistance, we must analyze the circuit in terms of the voltage across the external resistance and the total resistance in the circuit.

First, consider the two batteries in parallel:

• Battery 1: Voltage, \(V_1 = 10\,V\); internal resistance, \(r_1 = 0.5\,\Omega\).
• Battery 2: Voltage, \(V_2 = 12\,V\); internal resistance, \(r_2 = 0.8\,\Omega\).

The equivalent voltage \(V_{eq}\) and equivalent internal resistance \(R_{eq}\) for parallel batteries are given by:

\(V_{eq} = \frac{V_1/r_1 + V_2/r_2}{1/r_1 + 1/r_2}\)

Substitute the given values:

\(V_{eq} = \frac{10/0.5 + 12/0.8}{1/0.5 + 1/0.8} = \frac{20 + 15}{2 + 1.25} = \frac{35}{3.25} \approx 10.77\,V\)

For the equivalent internal resistance:

\(R_{eq} = \frac{(r_1 \cdot r_2)}{r_1 + r_2} = \frac{(0.5 \times 0.8)}{0.5 + 0.8} = \frac{0.4}{1.3} \approx 0.3077\,\Omega\)

Now, the total resistance, \(R_t\), in the circuit is the sum of \(R_{eq}\) and the external resistance, 20 \( \Omega \):

\(R_t = R_{eq} + 20 = 0.3077 + 20 \approx 20.3077\,\Omega\)

The current \(I\) through the 20 \( \Omega \) resistance can then be found using Ohm's law:

\(I = \frac{V_{eq}}{R_t} = \frac{10.77}{20.3077} \approx 0.53\,A\)

Therefore, the current flowing through the 20 Ω resistance is approximately 0.53 A.

Answer 2

To solve this, we first calculate the equivalent EMF of the parallel combination of the two batteries using the formula: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \] Then, we calculate the total current in the circuit, and use Ohm's Law: \[ I = \frac{V_{\text{eq}}}{R_{\text{total}}} \] Where \( V_{\text{eq}} \) is the equivalent EMF, and \( R_{\text{total}} \) is the total resistance, which includes the internal resistance of the batteries and the external resistance.