# 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**