Advertisements
Advertisements
प्रश्न
Two bulbs are marked 100 W, 220 V and 60 W, 110 V. Calculate the ratio of their resistances.
Advertisements
उत्तर
Given Ist Bulb
P1 = 100 W
V1 = 220 V
IInd Bulb
P2 = 60 W
V2 = 110V
We know P = `"V"^2/"R"`
So R = `"V"^2/"P"`
So `"R"_1/"R"_2 = ("V"_1^2/"P"_1)/("V"_2^2/"P"_2)`
`= ("V"_1^2 xx "P"_2)/("P"_1 xx "V"_2^2)`
`= (220^2 xx 60)/(100 xx 110 xx 110) = 2.4`
APPEARS IN
संबंधित प्रश्न
A battery of emf 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series.
Find:
1) Current in the circuit
2) The terminal voltage of the cell
3) The potential difference across 6Ω Resistor
4) Electrical energy spent per minute in the 4Ω resistor.
State whether a voltmeter has a high resistance of a low resistance. Give reason for your answer.
The p.d. across a lamp is 12 V. How many joules of electrical energy are changed into heat and light when:
a charge of 5 C passes through it?
The other name of potential difference is:
The V-I graph for a series combination and for a parallel combination of two resistors is shown in Fig – 8.38. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.
The device used for measuring potential difference is
Calculate the current flowing through each of the resistors A and B in the circuit shown in the following figure.
Write an expression for the electrical power spent in flow of current through a conductor in terms of current and resistance.
A negative charge will move from ______ to ______ potential.
