**Electric Current Exam2 and Problem Solutions**

**1. **Find the ratio of magnitude of two resistance made of same matter R_{1}/R_{2}?

**R _{1}=ρ.3L/(π.r^{2})**

**R _{2}=ρ.L/(π.4r^{2})**

**R _{1}/R_{2}=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;**

**V _{DC}=-ε'-(Ir')=-8-(3.2)=-8-6=-14 Volt **

**Potential between points A and B;**

**V _{AB}=ε-(I.r)=44-(3.2)**

**V _{AB}=38 Volt**

**3.** Find potential difference between points X and Y.

**V _{AB}=V_{B}-V_{A}=Σε-ΣR.I**

**V _{AB}=V_{B}-V_{A}=ε-ε'-(I.R+I.r')**

**V _{AB}=50-30-2(6+4)**

**V _{AB}=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.**

**I _{1}=V/3R, I_{2}=V/3R, I_{3}=V/3R,**

**I _{1}=I_{2}=I_{3}**

**W=I ^{2}.R.t**

**W _{1}=W_{2}=W_{3} and heat produced ;**

**Q _{1}=Q_{2}=Q_{3}**

**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;
I _{A}=3ε/R**

**I _{B}=2ε/R**

**I _{C}=ε/R**

**I _{A}>I_{B}>I_{C}**

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

**B _{A}>B_{B}>B_{C}**

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