15 A battery of three cells in series, each of e.m.f. 2 V and internal resistance 0.5 12 is connected to a 2 12 resistor in series with a parallel combination of two 3 12 resistors. Draw the circuit diagram and calculate: a. the effective external resistance b. the current in the circuit C. the lost volts in the battery d the current in one of the 3 12 resistors.
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Explanation:
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
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