Answer to Question #208182 in Electric Circuits for Mina

Question

When resistance of 2 ohm is connected across the terminals of the cell the current is 0.5 A and when the resistance is increased to 5 ohm the current is 0.25 A. The emf of the cell is ?

Answer 1

Gives

Resistance

$R=2\Omega$

Current

$I=0.5A$

Electromotive force

Emf=$I(r+R)\rightarrow(1)$

Emf of the cell=V

$V=I(r+R)$

Pu value

V=0.5(2+r)$\rightarrow(2)$

When resistance increase

R=5$\Omega$

$I=0.25A$

V=0.25(5+r)$\rightarrow(3)$

equation (2)and(3)are equal

Above both term potential is equal

0.25(5+r)=0.5(2+r)

(5+r)=2(2+r)

r=1$\Omega$

We know that

V=0.25(5+r)

Put r value

V=0.25(5+1)=1.5V

e.m.fof thr cell

V=1.5V