Question

A battery of e.m.f. 15V and internal resistance 3 Ω is
connected to two resistors 3 Ω and 6 Ω connected in parallel.
Find:–
• The current through the battery
• The p.d. between the terminals of the battery,
• The current in 3 Ω resistor,
• The current in 6 Ω resistor.

____ ___ ___
____ ____
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Answer 1

Explanation:

E= Ir

E=I(3+(3×6/3+6))

E=I(5)

E=5I

.I= E/5=3A

terminal potential

E=V+v

V= terminal potential

v= p.d across internal resistance

15= V+(3×3)

V= 6V

.the 3, 6 ohm are in parallel

so p.d or terminal potential difference(V) will be same

let current through 3 ohm be I1

V=3.I1

6= 3.I1

I1= 2A

and let current through 6 ohm be I2

V= 6I2

I2= 1A

Answer 2

Question :

A battery of e.m.f. 15V and internal resistance 3 Ω is

connected to two resistors 3 Ω and 6 Ω connected in parallel.

Find:–

• The current through the battery

• The p.d. between the terminals of the battery,

• The current in 3 Ω resistor,

• The current in 6 Ω resistor.

Answer;

• The current through the battery

→ in parallel 1/R = 1/3 + 1/6=1/2

so ,

R = 2 ohm

r=3 W

ε = 15 v

ε = 1 (R+ r)

15 = 1(2+3)

1=15/5

= 3 A

•The p.d. between the terminals of the battery

→ V = ?

R = 2 ohm

V = IR = 3x2 = 6v

•The current in 3 Ω resistor

→ V =6V

R= 3 ohm

I = v/ R = 6/3 = 2 A

•The current in 6 Ω resistor.

→ R= 6 ohm

v= 6v

I = v/R = 6/6 =1A