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.

Answer 3

This circuit has two batteries connected in parallel, both feeding a single external resistor of 20 ohms. Instead of using the shortcut formulas for equivalent EMF and equivalent internal resistance, the same result can be found by writing a direct node-voltage (Kirchhoff current law) equation at the junction where both batteries meet the external resistor. Let \( V \) be the voltage at that junction relative to the common negative terminal. The current leaving battery 1 into the node is \( \frac{10-V}{0.5} \), the current leaving battery 2 into the node is \( \frac{12-V}{0.8} \), and the current leaving the node through the external resistor is \( \frac{V}{20} \). Since current into the node equals current out:

\[ \frac{10-V}{0.5} + \frac{12-V}{0.8} = \frac{V}{20} \]

Multiplying out and collecting terms gives \( 35 - 3.25V = \frac{V}{20} \), and solving this for \( V \) gives \( V \approx 10.61 \, \text{V} \). The current through the 20 ohm resistor is then \( I = \frac{V}{20} \approx 0.53 \, \text{A} \). Let's check this against each option.

1. 0.75 A: This would need a junction voltage of \( 0.75 \times 20 = 15 \, \text{V} \), which is higher than either battery's own terminal voltage once its internal drop is accounted for, so this is too large.
2. 1.74 A: This would require a junction voltage of about \( 34.8 \, \text{V} \), far higher than either battery's EMF, which is not physically possible here.
3. 0.53 A: This matches the junction voltage of about \( 10.61 \, \text{V} \) found directly from the node equation, consistent with both batteries pushing current into the same node.
4. 1.21 A: This would need a junction voltage of about \( 24.2 \, \text{V} \), again higher than either EMF, so it cannot be right.

Solving the node equation directly, without relying on the equivalent-EMF formula, confirms the same junction voltage and current found by that shortcut.

Therefore, the correct answer is 0.53 A.