Answer to Question #79955 in Electric Circuits for Mission

Question

When a particular cell is in an open circuit,the pd between its terminal is 1.50v.when a 5ohm resistor is connected between the terminals,the pd drop to 1.25v .when the 5ohm resistor is replaced by the a resistance x,the pd become 1.00v .caculate (1) the internal resistance of the cell end (2) the value of the resistance x?

Answer 1

The Ohm’s law states
I=ℇ/(R+r),I=V/R
Where
ℇ=1.50 V is EMF of the battery
R=5 Ω is the external resistance
r is the internal resistance
V is the potential difference
Thus
V/R=ℇ/(R+r)

r=ℇ/V R-R=R(ℇ/V-1)=5×(1.50/1.25-1)=1 Ω

V/x=ℇ/(x+r)
Vx+Vr=ℇx
x=V/(ℇ-V) r=1.00/(1.50-1.00)×1=2 Ω