Question

(a) Two cells, each of emf 1.5 volt and internal
resistance 2 ohm are connected in series. Find the current when the cells are connected by a 1 ohm
resistor. [Ans. 0.6A]
(b) A cell having an emf of 1.5 volt and internal resistance 1 ohm is connected to
two resistances 2 and 3 ohms in series. Find the current flowing and the p.d. across the ends of each
resistance. (Ans. 0.25 A ; 0.5V, 0.75V]
(c) A cell supplies a current of 0.6A through a 2 ohm coil and a
current of 0.2A through a 7 ohm coil. Calculate the emf and internal resistance of the cell. (Ans. 1.5;0.5ohm]
​

Answer 1

Explanation:

a) E=nE=2×1.5V=3V

r=nr=2×2=4ohm

I=E/(r+R)=3/(4+1)=3/5=0.6 A

b) total external resistance=2+3=5ohm

I=E/(r+R)=1.5/(1+5)=1.5/6=0.25 A

P.D across 2ohm resistor= 2×0.25=0.5V

P.D across 3ohm resistor= 3×0.25=0.75V

c) I=E/(r+R)

0.6(r+2)=E

1.2=E-0.6r

similarly, 0.2(r+7)=E

1.4=E-0.2r

Subtracting 1st eq from 2nd,we get

0.2=0.4r

0.2/0.4=r

0.5ohm=r

E=I(R+r)

E=0.6(2+0.5)

E=0.6×2.5

E=1.5V