Question

A battery of e.m.f 20V and the internal resistance 0.2Ωsupplies a load taking 10A. Determine the p.d at the battery terminals of the load and the resistance of the load

Answer 1

Explanation

• Actual batteries have an internal resistance which drops some voltage at them reducing the sensible potential difference between the battery terminals that applies to any load connected to battery.
• In such cases sensible p.d (V) is given by E - i.r where E= e.m.f of the battery, i=current drawn from the battery, r= internal resistance.

Calculations

1). If the p.d is V then

$\qquad \begin{aligned} \small V &= \small E - ir\\ \small &= \small 20v-10A\times0.2\Omega\\ &= \small \bold{18v} \end{aligned}$

2). If the resistance of the load is R then,

$\qquad \begin{aligned} \small V &= \small iR\\ \small R &= \small \frac{18v}{10A}\\ &= \small \bold{1.8\Omega} \end{aligned}$

or applying Kirchhoff's laws to the entire circuit,

$\qquad \begin{aligned} \small E-ir-iR = 0\\ \small R &= \small \frac{E}{i}-r\\ &= \small \bold{1.8\Omega} \end{aligned}$