Equivalent EMF of Battery Combinations

This problem requires us to determine the equivalent electromotive force (EMF) of a circuit comprising multiple batteries connected in both series and parallel configurations. We will first analyze the series connection and then the parallel connection to find the final equivalent EMF.

Battery Series Connection Analysis

First, let's consider batteries E1 and E2 connected in series.

• Battery E1: EMF (\(E_1\)) = 3 V, internal resistance (\(R_1\)) = 0.5 Ω
• Battery E2: EMF (\(E_2\)) = 6 V, internal resistance (\(R_2\)) = 1.0 Ω

The problem states that the positive terminal of E2 is connected to the negative terminal of E1. This specific arrangement means that the batteries are connected in a series-aiding configuration, where their EMFs add up. When batteries are connected in series, their internal resistances also add up.

The equivalent EMF (\(E_{12}\)) of the series combination of E1 and E2 is:

$$E_{12} = E_1 + E_2$$ $$E_{12} = 3 \text{ V} + 6 \text{ V}$$ $$E_{12} = 9 \text{ V}$$

The equivalent internal resistance (\(R_{12}\)) of the series combination of E1 and E2 is:

$$R_{12} = R_1 + R_2$$ $$R_{12} = 0.5 \text{ Ω} + 1.0 \text{ Ω}$$ $$R_{12} = 1.5 \text{ Ω}$$

Battery Parallel Connection Analysis

Next, the third battery E3 is connected in parallel with the E1-E2 series combination.

• Combined battery (from E1 and E2): EMF (\(E_{12}\)) = 9 V, internal resistance (\(R_{12}\)) = 1.5 Ω
• Battery E3: EMF (\(E_3\)) = 6 V, internal resistance (\(R_3\)) = 1.0 Ω

For batteries connected in parallel, with EMFs \(E_a\), \(E_b\) and internal resistances \(r_a\), \(r_b\), the equivalent internal resistance (\(r_{eq}\)) and equivalent EMF (\(E_{eq}\)) are given by the following formulas:

Equivalent internal resistance:

$$ \frac{1}{r_{eq}} = \frac{1}{r_a} + \frac{1}{r_b} $$

Equivalent EMF:

$$ \frac{E_{eq}}{r_{eq}} = \frac{E_a}{r_a} + \frac{E_b}{r_b} $$

Let's calculate the equivalent internal resistance for the parallel combination of the (E1, E2) series combination and E3.

$$ \frac{1}{R_{eq}} = \frac{1}{R_{12}} + \frac{1}{R_3} $$ $$ \frac{1}{R_{eq}} = \frac{1}{1.5 \text{ Ω}} + \frac{1}{1.0 \text{ Ω}} $$ $$ \frac{1}{R_{eq}} = \frac{1}{3/2} + 1 $$ $$ \frac{1}{R_{eq}} = \frac{2}{3} + 1 $$ $$ \frac{1}{R_{eq}} = \frac{2 + 3}{3} $$ $$ \frac{1}{R_{eq}} = \frac{5}{3} \text{ Ω}^{-1} $$ $$ R_{eq} = \frac{3}{5} \text{ Ω} $$ $$ R_{eq} = 0.6 \text{ Ω} $$

Now, let's calculate the equivalent EMF (\(E_{eq}\)) for the parallel combination.

$$ \frac{E_{eq}}{R_{eq}} = \frac{E_{12}}{R_{12}} + \frac{E_3}{R_3} $$ $$ \frac{E_{eq}}{0.6 \text{ Ω}} = \frac{9 \text{ V}}{1.5 \text{ Ω}} + \frac{6 \text{ V}}{1.0 \text{ Ω}} $$ $$ \frac{E_{eq}}{0.6 \text{ Ω}} = 6 \text{ A} + 6 \text{ A} $$ $$ \frac{E_{eq}}{0.6 \text{ Ω}} = 12 \text{ A} $$ $$ E_{eq} = 12 \text{ A} \times 0.6 \text{ Ω} $$ $$ E_{eq} = 7.2 \text{ V} $$

Final Equivalent EMF

The equivalent EMF of the entire combination of batteries is 7.2 V.