Question

A battery of three cells in series ,each of e.m.f 2volts and internal resistance 0.5 ohms is connected to a 2 ohms resistor in series with a parallel combination of two 3 ohms 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 ohms resistors

Answer 1

Gives

(a)

Total resistance

$R=2+\frac{3\times3}{3+3}=3.5ohm\\R_t=0.5\times3+2+1.5=5ohm$

(b)

The external circuit reduced the battery voltage

Current in circuit

$I=\frac{V}{R_t}=\frac{6}{5}=1.2A$

(C)

Voltage drop across internal resistance

$V_i=IR_i$

$V_i=1.2\times0.5=0.6volt$

All three resistance voltage drop

$V_i=3\times0.5\times1.2=1.8V$

(d)

The current divides equally between the two 3 ohm resistor

$I'=\frac{I}{2}=\frac{1.2}{2}=0.6A$

This current through each 3 ohm resistor