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

When two batteries are connected in parallel, the equivalent electromotive force \( E_{\text{eq}} \) is determined by the following formula:

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

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

Substituting these values into the equation yields:

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

The combined equivalent emf of the batteries is \( 4 \, \text{V} \).

(ii): Calculating the Equivalent Internal Resistance

The equivalent internal resistance \( r_{\text{eq}} \) of the parallel combination is calculated using:

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

With the given values, the calculation is:

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

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

(iii): Calculating the Total Current

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

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

Ohm's law is applied to determine the current \( I \) drawn from the battery combination:

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

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