Advertisements
Advertisements
Question
An auto lamp is joined to a battery of e.m.f. 4 V and internal resistance 2.5Ω. A steady current of 0.5 A flows through the circuit. Calculate the
(a) Total energy provided by battery in 10 minutes,
(b) Heat dissipated in the bulb in 10 minutes.
Advertisements
Solution
Given, emf, e = 4 V, internal resistance r = 2.5Ω , current I= 0.5A
(a) Energy provided by battery in 10 mins =Power x time= (VIt)= 4 x 0.5 x 20= 40 watt-hour
(b) Heat dissipated in the bulb in 10 minutes= I2Rt
Let R be the resistance of the bulb, then:
I = `"e"/("R" + "r")`
or, 0.5 = `4/("R" + 2.5)`
or, R + 2.5 = 8
or, R = 5.5 Ω
Heat dissipated in the bulb in 10 minutes= I2Rt = ( 0.5)2 ( 5.5) (10 x 60) = 825 joules
APPEARS IN
RELATED QUESTIONS
If the charge on an electron is 1.6 × 10−19 coulombs, how many electrons should pass through a conductor in 1 second to constitute 1 ampere current?
Which effect of current is utilised in an electric light bulb?
Name two devices which work on the heating effect of electric current.
A wire of uniform thickness with a resistance of 27Ω is cut into three equal pieces and they are joined in parallel. Find the equivalent resistance of the parallel combination.
A cell supplies a current of 1.2 A through two 2 Ω resistors connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate:
(i) the internal resistance and (ii) e.m.f. of the cell.
What should I choose ___________________
b. for getting protection from electric current.
c. to measure the current in the circuit?
Electric current passes through a metallic conductor due to the movement of ______.
A continuous and closed path of an electric current is called:
