Name: Electrical Resistance and Resistivity
Uploaded: 2019-07-19
Duration: 2 min 37 s
Description: We have learnt that flow of charge constitutes an electric current. But, is the phenomenon of electric flow free to take place in the mediums? Let’s explore!

Question

Find the expression for the potential difference across the arrangement of the two cells in parallel in terms of their emf, internal resistance and total current in the given arrangement.

Accepted Answer

The parallel combination of cells is shown in the above figure. There are two cells of emf $ε_{1} and ε_{2}$ respectively. The total current is divided into a parallel combination of cells. The potential difference across the two cells will be:

$V_{1} = ε_{1} - I_{1} r_{1}$ .....(i)

$V_{2} = ε_{2} - I_{2} r_{2}$ .....(ii)

$r_{1} and r_{2}$ are internal resistances of the two cells.

On using (i) and (ii), the total current:

$I = I_{1} + I_{2} = \frac{ε_{1} - V}{r_{1}} + \frac{ε_{2} - V}{r_{2}} = \frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}} - V (\frac{1}{r_{1}} + \frac{1}{r_{2}})$ ......(iii)

The internal resistances of the two cells are in parallel, we can write for the equivalent resistance:

$\frac{1}{r_{e q}} = \frac{1}{r_{1}} + \frac{1}{r_{2}}$ (iv)

Using(iii) and (iv):

$\Rightarrow \frac{V}{r_{e q}} = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} r_{2}} - I$

$\Rightarrow V = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}} - \frac{I r_{1} r_{2}}{r_{1} + r_{2}}$ .........(v)

The above equation(v) gives the new potential difference across the parallel combination of cells.

David Sang and Darrell Hamilton Solutions for Chapter: Electrical Quantities, Exercise 1: Exercise 18.1

David Sang Physics Solutions for Exercise - David Sang and Darrell Hamilton Solutions for Chapter: Electrical Quantities, Exercise 1: Exercise 18.1

Questions from David Sang and Darrell Hamilton Solutions for Chapter: Electrical Quantities, Exercise 1: Exercise 18.1 with Hints & Solutions

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