A battery of three cells in series ,each of e.m.f. 2v and internal resistance 0.5 ohms is connected to a 2 ohm resistor in series with a parallel combination of two 3 ohms resistor . Draw the circuit diagram and calculate:

Question

a) the effective external resistance
b) the current in the circuit
c) the volts in battery
d) the current in one of the 3 ohm resistor

Answer 1

Effective resistance=R
Total resistance of circuit will be
R=2ohm +3×3\3+3ohm= (2+1.5)ohm=3.5
Equivalent resistance of the circuit Req= (3.5+1.5)=5ohm
Emf cell
V=E1+E2+E3=(2+2+2)v=6v
2. Current the circuit
=E/Req=6/5A
3. Lost voltage=v=Ir=6/5×1.5=1.8v
4. Current in the 3ohm resistor
(I1) =6/5×2=0.6A.
Hope it was helpful

Answer 2

Yes