Question

A battery of three cells in series, each of e.m.f 2v and internal resistance 0.5ohms is connected to a 2ohms resistor in series with a parallel combination of two 3ohms resistors. Draw the circuit diagram and calculate A. The effective resistance B. The current in the circuit. C. The lost volts in the battery. D. The current in one of the resistor.

Answer 1

I can't draw the circuit here, but i can analize the circuit:

a. Rt = 0.5 + 2 + (3 * 3) / (3 + 3) = 0.5 + 2 + 1.5 = 4 ohms = total

resistance in circuit.

b. The external circuit reduces the battery voltage.

c. I = V/Rt = 6 / 4 = 1.5Amps = the

current in circuit.

d. Vi = I*Ri = 1.5 * 0.5 = 0.75 volts

= Voltage drop across internal resistance.

e. The current divides equally between

the two 3 ohm resistors:

I3 = I / 2 = 1.5 / 2 = 0.75 Amps =

current through each 3 ohm resistor.

HOPE IT WILL HELP YOU SISTER

PLEASE MARK IT AS BRAINLIEST

Answer 2

Answer:

R=3.5ohms

I=1.2amps

v=1.8Volts

I³=0.6amps