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Question: How are cells combined in parallel? Derive the expression for the current flowing in the external ci...

How are cells combined in parallel? Derive the expression for the current flowing in the external circuit. When is this combination useful?

Explanation

Solution

When cells are connected in parallel, the current is divided among various cells. In a parallel combination, all the positive ends are connected together and all the negative ends are connected together.

Formula used:

In this solution, we’ll be using the following formula:

Ohm’s law: V=IRV = IR

where VV is the EMF of the cell, II is the current in the circuit which has resistance RR.

Complete step by step solution:

When cells are connected in parallel, the EMF of the combination will be equal to the EMF of any one of the cells that are connected in parallel. Let us assume that nn cells of emf EE are connected in parallel and have internal resistance rr.Since we know that the net EMF of all the cells in parallel is EE , let us now calculate the net internal resistance of all the cells. Since all the cells are combined in parallel, the internal resistances will be in parallel and we can calculate the net resistance as,

1rnet=1r+1r+....ntimes\Rightarrow \dfrac{1}{{{r_{net}}}} = \dfrac{1}{r} + \dfrac{1}{r} + ....{\text{n}}\,{\text{times}}

rnet=rn\Rightarrow {r_{net}} = \dfrac{r}{n}

Hence the net internal resistance will be rnet{r_{net}}.The total current in the circuit is given as,

Total current I=Total emftotal resistance\text{Total current I} = \dfrac{\text{Total emf}}{\text{total resistance}}

Here the total resistance is the sum of net internal resistance and external resistance, say RR.So the total resistance is rn+R\dfrac{r}{n} + R.

Since the potential/EMFs of the combination of all the cells is EE and we calculated the total resistance so the current in the circuit is given as,

I=nEr+nR\Rightarrow I = \dfrac{nE}{r+nR}

If r>nRr > nR then we obtain the expression of current as,

I=nEr\therefore I = \dfrac{{nE}}{r}

The current in the circuit is hence nn times the current in the circuit if only one cell were present. This is because the internal resistance decreases when the cells are attached in parallel. This is advantageous to us in situations when the internal resistance of the cell is very high and if we want to reduce the internal resistance, we can attach these cells in parallel and hence reduce the internal resistance of the circuit.

Note:

If we also have an external resistance in the circuit of resistance RR , it will be connected in series with the internal resistances and the net resistance will be rnet=rn+R{r_{net}} = \dfrac{r}{n} + R and then the current can be calculated from ohm’s law again as:

I=Er/n+R\Rightarrow I = \dfrac{E}{{r/n + R}} .