Q.An electrical circuit is shown below (a network with two branches, each carrying EMF ε1, internal resistance r1 with current I1, and EMF ε2, internal resistance r2 with current I2, joining at two nodes with terminal current I and terminal potential difference V). Show that V = (ε1 r2 + ε2 r1)/(r1 + r2) − I r1 r2/(r1 + r2).

Total emf of the 5 cells in series is 1 V and total internal resistance is 5 Ω; with the 10 Ω external resistor, current = 1/15 A.

For $n$ identical cells (emf $\varepsilon$, internal resistance $r$) connected in series with an external resistance $R$:

$I = \dfrac{n\varepsilon}{nr + R}$

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The two cells are in series, so their e.m.f.s and internal resistances add; applying $\varepsilon_{total} = I(R + r_{total})$ gives $R = 6\,\Omega$.

Setting up the circuit equation

From the figure, the two cells (each e.m.f. $\varepsilon = 2$ V, internal resistance $r = 1\,\Omega$) are connected in series with each other and with the external resistor $R$, all in one loop carrying current $I = 0.5$ A.

Since the cells are in series and aiding each other, the total e.m.f. of the combination is

$\varepsilon_{total} = 2 + 2 = 4\ \text{V}$ …

With no external load connected, the potential difference across A-B is simply the effective (Millman) EMF of the two parallel cells, which works out to 14 V.

The two branches (each an EMF in series with its internal resistance) are connected in parallel between the same two terminals A and B, with no external resistor drawing current from those terminals.

For two cells of EMF $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ connected in parallel (same polarity), the equivalent EMF between the terminals is

$\varepsilon_{eq} = \dfrac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1+r_2}$

Substituting $\varepsilon_1=18\ \text{V}, r_1=2\,\Omega, \varepsilon_2=12\ \text{V}, r_2=1\,\Omega$: …

Applying Kirchhoff's voltage law to the single loop (two cells in series aiding, in series with R) gives $R=6\,\Omega$.

Solution:

From the figure: the loop contains a resistor $R$ on the top arm and two cells, each of e.m.f. $2\text{ V}$ and internal resistance $1\,\Omega$, connected in series (one after another, same polarity) along the bottom arm, with plain connecting wire on the other side. The whole loop carries a single current $I = 0.5\text{ A}$.

Total e.m.f. of the two series cells:

$\varepsilon = 2\text{ V} + 2\text{ V} = 4\text{ V}$

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Cells of equal EMF in parallel keep the EMF unchanged (only current capacity/effective internal resistance changes). So EMF $=6$ V.

When $n$ identical cells each of EMF $E$ are joined in parallel, the combined EMF is still $E$ (they behave like one cell that can supply more current). The in …

For two cells connected in parallel, the combination behaves like a single cell whose emf is a resistance-weighted average of the two individual emfs, and whose internal resistance is the parallel combination of the two internal resistances.

For two cells of emf $\varepsilon_1, \varepsilon_2$ and internal resistances $r_1, r_2$ connected in parallel (like terminals joined together) and delivering current $I$ to an external circuit, applying Kirchhoff's laws to the two branches and equating terminal voltages gives th …

Equivalent emf of two parallel cells: $E_{eq}=\dfrac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}=5\,V$, option C..

For two cells of emf $E_1,E_2$ and internal resistances $r_1,r_2$ connected in parallel with like terminals together, the combination is equivalent to a single cell of emf $E_{eq}$ and internal resistance $r_{eq}$, where

$\frac{1}{r_{eq}} = \frac{1}{r_1}+\frac{1}{r_2}, \qquad \frac{E_{eq}}{r_{eq}} = \frac{E_1}{r_1}+\frac{E_2}{r_2}$

Given $E_1=6\,V,\ r_1=1\,\Omega$; $E_2=3\,V,\ r_2=2\,\Omega$:

$r_{eq} = \frac{r_1r_2}{r_1+r_2} = \frac{1\times2}{1+2} = \frac23\,\Omega$

$E_{eq} = r_{eq}\left(\frac{E_1}{r_1}+\frac{E_2}{r_2}\right) = \frac23\left(6+1.5\right) = \frac23\times7.5 = 5\,V$

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