Question

1. Two electric bulbs are marked 60W,220V and 60W, 110V repectively calculate the ratio of their resistence

2. A bulb is marked 10V,40W operates on a 10V battery for 5 minutes. Calculate :
a. current flowing through it
b. energy liberated

3. An electric kettle is rated 1.5 kW, 250V. Find the cost of running the kettle for 3 hours at Rs.250 per unit

4. Three equal resistances connected in series take 10 W power from the main supply. If these resistances be connected in parallel, then how much power will they take.

5. Water in an electric kettle connected at a 240 V supply took 4 minutes to reach its boiling point. How long would it have taken if the supply has been of 200 V.

Answer 1

1)Formula for calculating Power P is P=V*I or P=V^2/R

As we can see that V is same for both the bulbs. So, by considering the above formula P is inversely proportional to R(Resistance).So, bulb giving power of 60 watt will hae higher resistance.

Hope it helps!

4)When the three resistors are in series, effective resistance in the circuit = 3R.
Power dissipated in the series circuit = Pseries = V2/(3R) = 10 W
                                                                          ⇒ V2/(R) = 30 W
When the three resistors are in parallel, effective resistance in the circuit = R/3.
Power dissipated in the parallel circuit = Pparallel = V2/(R/3) = 3V2/(R) 

Answer 2

Hi..

Hope this helps u!!