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GE6252-Basic-Electrical-and-Electronics-Engineering-Anna-university-notes-->View question

Asked On2017-06-20 12:48:20 by:milan-ransingh

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The resistance of electric lamp, R1 = 20 Ω,
The resistance of the conductor connected in series, R2 = 4 Ω.
Then the total resistance in the circuit
R =R1 + R2
Rs = 20 Ω + 4 Ω = 24 Ω.
The total potential difference across the two terminals of the battery
V = 6 V.
Now by Ohm’s law, the current through the circuit is given by
I = V/Rs
= 6 V/24 Ω
= 0.25 A.
Applying Ohm’s law to the electric lamp and conductor separately, we get potential difference across the electric lamp,
V1 = 20 Ω × 0.25 A
= 5 V;
and, that across the conductor, V2 = 4 Ω × 0.25 A
= 1 V.
Suppose that we like to replace the series combination of electric lamp and conductor by a single and equivalent resistor. Its resistance must be such that a potential difference of 6 V across the battery terminals will cause a current of 0.25 A in the circuit. The resistance R of this equivalent resistor would be
R = V/I
= 6 V/ 0.25 A
= 24 Ω.
This is the total resistance of the series circuit; it is equal to the sum of the two resistances.

Answerd on:2017-06-20 Answerd By:Aparna-Dasgupta

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