# Two cells, each of e.m.f 1.5V and internal resistance 1Ω are connected in parallel, to form a battery.

+1 vote
16.6k views
in Physics

Two cells, each of e.m.f 1.5V and internal resistance 1Ω are connected in parallel, to form a battery. The battery is connected to an externari resistance of 0.5Ω and two resistances of 3Ω and 1.5Ω in parallel.

(a) Draw the circuit diagram.

(b) Calculate the current in main circuit.

(c) Calculate the current in 1.5Ω resistor.

(d) Calculate the drop in potential across the terminal of the battery.

by (72.6k points)
selected

(a) Circuit diagram (b) Current in the main circuit Total internal resistance of two parallel cells I/R1 = 1 + 1

R = 1/2 = 0.5Ω

Effective resistance between PQ = 1/R2 = 1/3 + 2/3

R2 = 1Ω

Total resistance of circuit R = R1 + R2+ external resistance

R = 0.5 + I + 0.5 = 2Ω, V= 1.5V

current in main circuit

I = V/R = 1.5/2 = 0.75A

(c) Current in 1 .5Ω resistor

p.d. between PQ = IR2

V1 = 0.75 × 1

V1 = 0.75V

V1 = 0.75

I1= V1 /1.5 = 0.75/1.5 = 0.5A

(d) Drop in potential across the terminals of battery E – V = Ir = 0.75 × 0.5 = 0.375V

+1 vote