Question

A battery'sopen circuit voltage is 2volts when a 500ohm resistor is connected to the battery ,voltage drops to 1volt.What is the internal resistance of battery

Answer 1

E.m.f of cell (e) =2V
Voltage drop(v) =1V
Therefore, terminal voltage(V) =2-1=1V
Now,
Resistance(R)=500 ohm
Therefore internal resistance(r) =[(e/V) - 1]*R
r=[(2/1)-1]*500
= (2-1)*500
=1*500
=500ohm

Answer 2

Given,

Voltage=2v

Ω= 500 0hm

To find,

the internal resistance of the battery

Solution,

The Thevenin equivalent happens to be the actual circuit the power dissipation of the series resistance is

Power of series resistance =1V^2/500 = 0.002 W

Power Load = 1V^2/500 = 0.002 W

Thus the internal resistance of the battery is  0.004 W.

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