Name: Applications of Potentiometer
Uploaded: 2019-07-19
Duration: 4 min 30 s
Description: Since potentiometers are very versatile and handy to use, it is vital to understand their working principle and applications fully. In this video, learn abou...

Question

Deduce an expression for the current supplied by a mixed grouping of cells to an external resistance

R

. Hence, obtain the condition for maximum current delivered by the combination.

Accepted Answer

Sometimes cells are grouped in series and in parallel to get a described current. It is particularly necessary when the internal resistance of the cells cannot be neglected in comparison with the external resistance $R_{e}$ . Let us take a combination in which all the cells have the same emf $ε$ and the same internal resistance $r$ . Let there be $m$ rows of cells connected in parallel with n cells in series in each rows as shown in figure below.

Obviously the effective emf of each row= $n ε = ε_{s}$ (say) since there are $n$ cells in series. Also the effective internal resistance of each row = $n \times r = r_{s}$ (say). As there are $m$ rows of cells in parallel, the effective emf $ε_{e}$ of the combination = the effective emf of each row. Thus

$ε_{e} = ε_{s} = n ε$

Similarly, the effective internal resistance $r_{e}$ of the combination is given by

$\frac{1}{r_{e}} = \frac{1}{r_{s}} + \frac{1}{r_{s}} + . . . . . . m times \Rightarrow \frac{1}{r_{e}} = \frac{m}{r_{s}} = \frac{m}{nr} \Rightarrow r_{e} = \frac{nr}{m}$

The total resistance of the circuit is given by

$R_{t} = R_{e} + r_{e} = R_{e} + \frac{n r}{m}$

Where $R_{e}$ stands for the external resistance. The current in the circuit is given by

$I = \frac{ε_{e}}{R_{t}} = \frac{n ε}{R_{e} + \frac{n r}{m}} \Rightarrow I = \frac{m n ε}{m R_{e} + n r}$

Condition for maximum current

Since the numerator in the R.H.S. of above equation is a constant, the current in the circuit becomes maximum when the denominator is minimum. We may write the denominator $m R_{e} + n r$ as follows.

$m R_{e} + n r = {(\sqrt{m R_{e}})}^{2} + {(\sqrt{n r})}^{2} = {(\sqrt{m R_{e}} - \sqrt{n r})}^{2} + 2 m n R_{e} r$

Since $2 \sqrt{m n R_{e} r}$ is a fixed quantity in a given combination, $(m R_{e} + n r)$ is minimum when ${(\sqrt{m R_{e}} - \sqrt{n r})}^{2} = 0$ . This leads to the condition that

$m R_{e} = n r \Rightarrow R_{e} = \frac{n r}{m} = r_{e}$

Thus, the maximum current is obtained for the circuit when the effective internal resistance of all the cells becomes equal to the effective external resistance. Such maximum current $I_{m a x}$ is given by

$I_{m a x} = \frac{m n ε}{2 \sqrt{m n R_{e} r}} = \frac{m n ε}{2 \sqrt{m R_{e}} . \sqrt{n r}}$

$I_{m a x} = \frac{m n ε}{2 m R_{e}} o r \frac{m n ε}{2 n r}$ [using $m R_{e} = n r$ from previous equation]

$I_{m a x} = \frac{n ε}{2 R_{e}} o r \frac{m ε}{2 r}$

Deduce an expression for the current supplied by a mixed grouping of cells to an external resistance $R$ . Hence, obtain the condition for maximum current delivered by the combination.

Important Questions on Direct Current Circuits

Learn and Prepare for any exam you want !

Deduce an expression for the current supplied by a mixed grouping of cells to an external resistance R. Hence, obtain the condition for maximum current delivered by the combination.

Important Questions on Direct Current Circuits

Learn and Prepare for any exam you want !

Deduce an expression for the current supplied by a mixed grouping of cells to an external resistance $R$ . Hence, obtain the condition for maximum current delivered by the combination.