An electric lamp, whose resistance is 20 `Omega `, and a conductor of 4 `Omega ` resistance are connected to a 6 V battery (Fig. 12.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Explanation

The resistance of electric lamp, `R_1 = 20 Omega `
The resistance of the conductor connected in series,`R_2 = 4 Omega `.
Then the total resistance in the circuit
`R = R_1 + R_2`
` R_s = 20 Omega + 4 Omega = 24 Omega `.
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//R_s`
`= 6 V//24 Omega `
` = 0.25 A.`