Question

Two batteries of emfs 3V and 6V and internal resistances 0.2 and 0.4 are connected in parallel. This combination is connected to a 4 resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination

Hint:

For parallel connections of batteries, the equivalent emf is a weighted average of the individual emfs, and the equivalent internal resistance is found using the formula for parallel resistances.

Answer 1

Battery Calculation: Equivalent Emf, Internal Resistance, and Current 

(i): Calculating the Equivalent Emf

The two batteries are connected in parallel, so the equivalent emf \( E_{\text{eq}} \) is calculated using the formula:

\[ E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \]

Given values: \( E_1 = 3 \, \text{V}, E_2 = 6 \, \text{V}, r_1 = 0.2 \, \Omega, r_2 = 0.4 \, \Omega \)

Substitute the values into the equation:

\[ E_{\text{eq}} = \frac{(3 \times 0.4) + (6 \times 0.2)}{0.2 + 0.4} = \frac{1.2 + 1.2}{0.6} = \frac{2.4}{0.6} = 4 \, \text{V} \]

Thus, the equivalent emf of the combination is \( 4 \, \text{V} \).

(ii): Calculating the Equivalent Internal Resistance

The equivalent internal resistance \( r_{\text{eq}} \) is given by:

\[ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} \]

Substitute the values:

\[ r_{\text{eq}} = \frac{(0.2 \times 0.4)}{0.2 + 0.4} = \frac{0.08}{0.6} = 0.1333 \, \Omega \]

Thus, the equivalent internal resistance of the combination is \( 0.1333 \, \Omega \).

(iii): Calculating the Total Current

The total resistance in the circuit is the sum of the internal resistance \( r_{\text{eq}} \) and the external resistor \( R = 4 \, \Omega \). Thus, the total resistance \( R_{\text{total}} \) is:

\[ R_{\text{total}} = r_{\text{eq}} + R = 0.1333 + 4 = 4.1333 \, \Omega \]

Using Ohm's law, the current \( I \) drawn from the combination is:

\[ I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{4}{4.1333} = 0.968 \, \text{A} \]

Thus, the current drawn from the combination is \( 0.968 \, \text{A} \).