Question

An electric lamp whose resistance is 20 ohm and and conductor of 4 ohm resistance are connected to a battery of 6 volt calculate(a) total resistance current through circuit
(B) the current through the circuit
(C) the potential difference across lamp (1) electric lamp and (2) conductor
(D) power lamp

Answer 1

Answer:

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