Question

A battery of three cells in series, each of emf 2V and internal resistance 0.5ohms, is connected to a 2ohms resistor in series with a parallel combination of two 3ohms resistors.Draw the diagram and calculate

1)the effective external resistance

2)the current in the circuit

3)the lost volts in the battery

4)the current in one of the 3ohms resistor.

Answer 1

The circuit diagram of the given question is

i) Let the effective external resistance is represented by $R$

total external resistance of the circuit will be $(R)=2\Omega +\frac{3\times 3}{3+3} \Omega=(2+1.5)\Omega=3.5\Omega$

ii) Equivalent resistance of the circuit $R_{eq}=(3.5+1.5)\Omega = 5\Omega$

net emf of the cell $V=E_1+E_2+E_3 = (2+2+2)V= 6V$

Hence, current in the circuit $(i)=\frac{E}{R_{eq}}$

$(i)=\frac{6}{5}A$

iii) The lost volts in the battery $v=ir = \frac{6}{5}\times 1.5 =1.8V$

iv) Current in the one of the $3\Omega$ resistance will be $(i_1)=\frac{6}{5\times 2}=0.6A$