Electric Current Exam 2 and Problem Solutions

1. Find the ratio of magnitude of two resistance made of same matter R₁/R₂? R₁=ρ.3L/(π.r²)

R₂=ρ.L/(π.4r²)

R₁/R₂=12

2. Find the potential difference between points A and B. Current passing through circuit is;

I=(ε-ε’)/(R+r+r’)=(44-8)/(8+2+2)

I=3 Amperes

Potential between points D and C;

VDC=-ε’-(Ir’)=-8-(3.2)=-8-6=-14 Volt

Potential between points A and B;

VAB=ε-(I.r)=44-(3.2)

VAB=38 Volt

3. Find potential difference between points X and Y. VAB=VB-VA=Σε-ΣR.I

VAB=VB-VA=ε-ε’-(I.R+I.r’)

VAB=50-30-2(6+4)

VAB=20-20=0

4. Find relation between heat produced by each branch of circuit given below. Since branches are in parallel potential differences between them are equal.

I₁=V/3R, I₂=V/3R, I₃=V/3R,

I₁=I₂=I₃

W=I².R.t

W₁=W₂=W₃ and heat produced ;

Q₁=Q₂=Q₃

5. If we close switches of circuits given below, find the changes in the brightness of the bulbs. (Neglect resistances of the batteries) When we close switches, currents on A, B and C are;

IA=3ε/R

IB=2ε/R

IC=ε/R

IA>IB>IC

Brightness of the bulbs is directly proportional to currents passing on them.

BA>BB>BC