Current Electricity - Combination of Cells in Series and Parallel

Two cells of EMF $2\text{ V}$ and $4\text{ V}$ with internal resistances $1\text{ }\Omega$ and $2\text{ }\Omega$ respectively are connected in parallel. Calculate the equivalent EMF and equivalent internal resistance of the combination.

Given: $E_1 = 2\text{ V}$, $r_1 = 1\text{ }\Omega$, $E_2 = 4\text{ V}$, $r_2 = 2\text{ }\Omega$. Using the parallel formula for equivalent EMF: $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$ $E_{eq} = \frac{(2 \times 2) + (4 \times 1)}{1 + 2} = \frac{4 + 4}{3} = \frac{8}{3}\text{ V} \approx 2.67\text{ V}$ Using the parallel formula for equivalent internal resistance: $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$ $r_{eq} = \frac{1 \times 2}{1 + 2} = \frac{2}{3}\text{ }\Omega \approx 0.67\text{ }\Omega$

In a parallel combination of non-identical cells, the equivalent EMF is a weighted average of the individual EMFs, weighted by the reciprocal of their internal resistances. The equivalent resistance follows the standard parallel law for resistors.

Four identical cells each of EMF $1.5\text{ V}$ and internal resistance $0.5\text{ }\Omega$ are connected in series to an external resistor of $8\text{ }\Omega$. Find the current flowing in the circuit.

Given: $n = 4$, $E = 1.5\text{ V}$, $r = 0.5\text{ }\Omega$, $R = 8\text{ }\Omega$. Total EMF $E_{total} = nE = 4 \times 1.5 = 6\text{ V}$. Total internal resistance $r_{total} = nr = 4 \times 0.5 = 2\text{ }\Omega$. Total circuit resistance $R_{total} = R + nr = 8 + 2 = 10\text{ }\Omega$. Current $I = \frac{E_{total}}{R_{total}}$: $I = \frac{6}{10} = 0.6\text{ A}$

When cells are in series, their EMFs and internal resistances are simply added. The current is then calculated using Ohm's law for the complete circuit including the external load.

Three identical cells, each of EMF $2\text{ V}$ and internal resistance $0.2\text{ }\Omega$, are connected in series. However, one cell is accidentally connected with reversed polarity. Calculate the net EMF and the current in the circuit if an external resistor of $5.4\text{ }\Omega$ is used.

1. Identify parameters: $n = 3$, $E = 2\text{ V}$, $r = 0.2\text{ }\Omega$, $R = 5.4\text{ }\Omega$.
2. Since one cell is reversed, net EMF $E_{net} = (n - 2n_{reversed})E = (3 - 2(1)) \times 2 = 1 \times 2 = 2\text{ V}$.
3. Total internal resistance $r_{total} = n \times r = 3 \times 0.2 = 0.6\text{ }\Omega$.
4. Total resistance $R_{total} = R + r_{total} = 5.4 + 0.6 = 6.0\text{ }\Omega$.
5. Current $I = \frac{E_{net}}{R_{total}} = \frac{2}{6} = 0.33\text{ A}$.

Even though a cell is reversed, its internal resistance still contributes to the total resistance of the circuit. The reversed EMF opposes the EMF of one of the correctly connected cells, effectively canceling it out.

Two cells of EMF $10\text{ V}$ and $2\text{ V}$ with internal resistances $2\text{ }\Omega$ and $1\text{ }\Omega$ respectively are connected in parallel with their like terminals together. Calculate the current through an external resistor of $4\text{ }\Omega$.

1. Calculate equivalent EMF $E_{eq}$: $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$ $E_{eq} = \frac{(10 \times 1) + (2 \times 2)}{2 + 1} = \frac{10 + 4}{3} = \frac{14}{3}\text{ V} \approx 4.67\text{ V}$
2. Calculate equivalent internal resistance $r_{eq}$: $r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\text{ }\Omega \approx 0.67\text{ }\Omega$
3. Total resistance of circuit $R_{tot} = R + r_{eq} = 4 + \frac{2}{3} = \frac{14}{3}\text{ }\Omega$.
4. Current $I = \frac{E_{eq}}{R_{tot}} = \frac{14/3}{14/3} = 1\text{ A}$.

When cells of different EMFs are in parallel, the equivalent EMF is a weighted average based on internal resistances. The equivalent resistance is found using the parallel resistor formula.