Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. How much current is drawn from the line if the supply voltage is 220 V? (Keep your answer to 2 decimal places)

Open in App

Solution

Resistance of the first lamp, R₁= $\frac{V^{2}}{P}$ = 484 Ω

Resistance of the first lamp, R₁= $\frac{V^{2}}{P}$ = 806.7 Ω

R_parallel= $\frac{484 * 806.7}{484 + 806.7}$ ) = 302.6 Ω

I = V/R_p= \(\frac{220}{302.6}\ = 0.73 A

Suggest Corrections

Similar questions

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Q. Question 15
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to an electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Q. Two lamps one rated

60 W

220 V

and the other

40 W

220 V

are connected in parallel to a

220 V

electric supply mains. What is the total current drawn from the electric mains if the voltage of the electric supply is

220 V

Q. Show how would you join three resistors, each of resistance 9

Ω

so that the equivalent resistance of the combination is (a) 13.5

Ω

(b) 6

Ω

?

OR

(a) Write Joule's law of heating.

(b) Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.

Q. A lamp rated

100 W

22 V

is connected to the main electric supply. What current is drawn from the supply line if the voltage is

220 V

Join BYJU'S Learning Program

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. How much current is drawn from the line if the supply voltage is 220 V? (Keep your answer to 2 decimal places)

Resistance of the first lamp, R1 = V2P = 484 Ω Resistance of the first lamp, R1 = V2P = 806.7 Ω Rparallel = 484∗806.7484+806.7) = 302.6 Ω I = V/Rp = \(\frac{220}{302.6}\ = 0.73 A

Resistance of the first lamp, R₁= $\frac{V^{2}}{P}$ = 484 Ω

Resistance of the first lamp, R₁= $\frac{V^{2}}{P}$ = 806.7 Ω

R_parallel= $\frac{484 * 806.7}{484 + 806.7}$ ) = 302.6 Ω

I = V/R_p= \(\frac{220}{302.6}\ = 0.73 A