Question

A battery of four cells each of e.m.f 1.5V and internal resistance 1.0ohms is connected to a 1ohm resistor in series with a parallel combination of two 2ohms resistors. Draw the circuit diagram and calculate:
a) the combined external resistance.
b) the current in the circuit
c) the lost volts in the battery
d) the current in one of the two resistors.

Answer 1

The circuit diagram has the following outline:

Here we have: $R_1=1 \, \Omega$ , $R_2=2 \, \Omega$ , $\Epsilon = 1.5 \,V$, $r = 1 \, \Omega$

1) the combined external resistance is

$R = R_1+\frac{R_2}{2} = 1 + 1 = 2 \, \Omega$

2) the current in the circuit is

$I = \frac{4 \Epsilon}{4r + R}=\frac{4 \cdot 1.5}{4+2}=1 \,A$

3) the lost volts in the battery are

$U=4 Ir=4 \cdot 1\cdot 1 = 4 \, V$

4) the current through $R_2$ is

$I_2 = \frac{I}{2} = 0.5 \,A$