Question

a piece of copper wire of resistance R is cut into 10 equal lengths. These parts are connected in parallel. How does the joint resistance of the parallel combination compare with the original resistance of the single wire?

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistance of 0.10 ohms and the voltmeter of 3.0 ohms; the ammeter reads 2.5 amps and the voltmeter reads 6.0 volts. What is the uncorrected value of the resistance in each case? What is the value of the resistance in each case?

Answer 1

originakl resistance R

when divided into 10 equal parts

each part = R/10

RNet in parallel = (1/Rnet) = 10./R + 10/R ..................

1/Rnet = 100/R

Rnet = R/100

so ratio of Rnet/R = 1/100 = 0.01

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uncorrected value of resistance= 6/2.5 = 2.4 ohm

this is after addition of 0.10 ohm in series and 3 ohms in parallel

so hence let correct value be x

x+0.10||3= 2.4

hence, x+0.10= 12

and x= 11.90 ohm