Let two resistors R1 and R2 are connected in parallel,

1/Req=1/R1+1/R2

Req=R1R2/(R1R2)

The equivalent resistance of a parallel combination is less than the resistance of each of the resistors.

You can use this result to explain why the resistance of a metal wire is inversely proportional to its area of cross section. Let us imagine a thick wire as a parallel combination of several thin wires. Then the resistance of the combination is less than that of each thin wire. In other words, the resistance of a thick wire is less than that of a thin wire.

Example 1: Three resisters 10 Ω, 20 Ω, 30 Ω, are connected in (a) Series (b) Parallel. Find the resultant resistance in the circuit. Solution: FromthegivencircuitR1 =10Ω,R2 =20Ω,R3 =30Ω
(a) Resultant resistance in Series connection: R = R1+ R2+ R3

R=10+20+30=60Ω

(B) Resultant resistance in parallel connection: 1/R=1/R1+1/R2+1/R3
1/R=1/10+1/20+1/30=1/60
R = 5.5 Ω

Example 2:Three resisters R1Ω, 4 Ω, and 8 Ω are connected in series in CERT TELANGAN a circuit. If the resultant resistance in the circuit is 20 Ω find the value of R1.
Solution: Resultant resistance in Series connection: R = R1+ R2+ R3
20 = R1+ 4 + 8
20=R1+12
R1 =20-12
R1 = 8 Ω
Example 3: Two resisters R1Ω, and 12 Ω, are connected in parallel in a circuit. If the resultant resistance in the circuit is 3Ω find the value of R1.
Resultaint resistance in parallel connection:1/R=1/R1+1/R2
1/3=1/R1+1/12
1/R1=1/3-1/12=3/12=1/4
R1=4Ω


page no:195

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