Answer to Question #140008 in Electric Circuits for Isaac Braimah

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

Total resistance of the circuit,

$R = (0.5 \times 3)+2+(\frac{3\times 3}{3+3}) = 5 \Omega$

Total potential of the circuit, $V = 3E = 3\times 2 = 6 V$

Current in the circuit, $I = \frac{V}{R} = \frac{6}{5} = 1.2 A$

Voltage lost in cell, $V' = E - Ir = 2 - (1.2 \times 0.5) = 1.4 V$

Current in combination of 3 ohm is 1.2 A. Since, both 3 ohm resistances are in parallel, so potential will be same. So, current in each of the resistance is half of the total resistance i.e. 0.6 A.