Question

An electric bulb is labeled 100W, 240V. Calculate (i) The current through the filament

(ii) The resistance of the filament used in the bulb​

Answer 1

Answer:

W=100, v=240 so we use W= V/Q

100 =240/Q

100Q = 240

Q = 2.4

And now we use V = IR

240 = 2.4R

R = 100

Answer 2

Answer:

An electric bulb labelled 100W, 240V. The current through the filament is 0.416A and resistance of the filament is 576.92 ohms.

Explanation:

• The rate of work done by and source of voltage, V in maintaining a current, I in the circuit is called power.
• If a current, I flow through a circuit for time t at a constant potential difference, V, then the work done is W = VIt
• Then, Electric power, P of circuit is

$P = \frac{W}{t} = VI$

• The SI unit of electric power is watt (W).

Given that :

• Power rating of bulb = 100 W
• Voltage = 240 V

Solution :

• Current through the filament is given by

$current \: I = \frac{P}{V} \\ = \frac{100}{240} \\ = 0.416 \: A$

• Resistance of the filament is given by

$Resistance \: R = \frac{V}{I} \\ = \frac{240}{0.416} \\ = 576.92 \: ohms$

• Hence, current through filament is 0.416 A and Resistance of filament is 576.92 ohms.