Since both the bulbs are connected in parallel, the voltage across each of them will be the same.
Current drawn by the bulb of rating 100 W can be calculated as follows:
\( P = V × I\)
\( I=\frac{P}{V}\)
Substituting the values in the equation, we get
\( I = \frac{100\,W}{220\,V}=\frac{100}{220\,A}\)
Similarly, the current drawn by the bulb of rating 60 W can be calculated as follows
\( I = \frac{60\,W}{220\,V}=\frac{60}{220\,A}\)
Therefore, the current drawn from the line is \( \frac{100}{220}+\frac{60}{220}= 0.727\,A\)
Answered by Shivani Kumari | 1 year agoExplain the following.
a. Why is the tungsten used almost exclusively for filament of electric lamps?
b. Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
c. Why is the series arrangement not used for domestic circuits?
d. How does the resistance of a wire vary with its area of cross-section?
e. Why copper and aluminum wires are usually employed for electricity transmission?
An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?