Question

A cell supplies 5600 J of energy when 4200 C of charge moves completely around the circuit. The load resistor is 7.5 ȍ and the potential difference across the load is 1.1 V. Find the emf, the current through the circuit, and the size of the internal resistance of the cell.  

Answer 1

Explanations & Calculations

• Start with the power that is supplied to the circuit from the battery.

$\qquad\qquad \begin{aligned} \small 5600&=\small E\cdot i\cdot t\cdots\cdots(1) \end{aligned}$

• Then replace time as follows,

$\qquad\qquad \begin{aligned} \small Q&=\small it\to t= \frac{Q}{i}\\ \small t&=\small \frac{4200\,C}{i}\\ \\ \small 5600&=\small E\cdot i\cdot \frac{4200}{i}\\ \small E&=\small \frac{4}{3}\,V \end{aligned}$

• From the details given for the load, the current can be calculated,

$\qquad\qquad \begin{aligned} \small V&=\small iR\\ \small 1.1&=\small i\cdot 7.5\Omega\\ \small i &=\small \frac{11}{75}\Omega \end{aligned}$

• Due to the internal resistance of the cell, its EMF is not applied to the load at it is, rather somewhat reduced. From that hint you can calculate the internal resistance.

$\qquad\qquad \begin{aligned} \small V_{applied}&=\small E-i\cdot r_{internal}\\ \small r_{internal}&=\small \frac{E-V_{applied}}{i} \end{aligned}$

• You can give it a try.