Question

6 identical cell each of emf 2v and internal resistance 1 are connected in parallel then the maximum power dissipated through external resistor of 10 is

Answer 1

Given:

6 identical cell each of emf 2v and internal resistance 1 are connected in parallel.

To find:

Max power dissipated through external resistance of 10 ohm?

Calculation:

First of all , let's calculate the equivalent potential difference provided by the combination of cells.

$\therefore \: E_{net} = \dfrac{ \frac{E1}{r1} + \frac{E2}{r2} + .... \: \frac{E6}{r6} }{ \frac{1}{r1} + \frac{1}{r2} + .... \: \frac{1}{r6} }$

$\implies\: E_{net} = \dfrac{ \frac{E}{r} + \frac{E}{r} + .... \: \frac{E}{r} }{ \frac{1}{r} + \frac{1}{r} + .... \: \frac{1}{r} }$

$\implies\: E_{net} = \dfrac{6 \times \frac{E}{r} }{ 6 \times \frac{1}{r} }$

$\implies\: E_{net} =E$

$\implies\: E_{net} =2 \: volt$

Net Internal resistance = 1/6 ohm.

$\therefore \: i = \dfrac{voltage}{resistance}$

$\implies \: i = \dfrac{2}{ \frac{1}{6} + 10}$

$\implies \: i = \dfrac{2}{ 0.16 + 10}$

$\implies \: i = \dfrac{2}{ 10.16}$

$\implies \: i = 0.19 \: amp$

So, current is 0.19 amperes