Physics Chapter 1
Effects of Electric current
2 Marks Expected Question
1. "Fuse wires of appropriate amperage should be selected" . Explain.
Answer : Current through the circuit varies in accordance to the power of devices in the circuit. This must be considered to select the amperage of fuse wire. In using a fuse wire of amperage less than what is necessary, it melts easily. In using a fuse wire of high amperage than what is necessary it doesn't blow out, even if the device damages.
2) a) 5 resistors of 5 Ω are qiven. What is the minimum resistance constructed by usinq these resistors?
b.) How many resistors of 10 Ω are needed to connect, in the same way, to qet the same effective resistance?
Answer : a) . 1 Ω ( b) .10 resistors.
3) How can the LED bulbs be disposed to scientifically?
Ans: Segregate the plastic, electronic and metal components of LED bulb and transfer them to their respective disposal units.
4) . In the following statements some of them are related to each other. Tabulate them.
a) Alloy of tin and lead
b) High melting point
c) When electric current is increased, fuse
d) Ability to emit white light.
Answer : a, c - Fuse wire and b, d - Filament of a bulb.
5a) Which are the factors that affect the heat produced in heating appliances?
b) How they can be related?
Ans: a) Intensity of current, resistance of the conductor and the time of flow of
current.
b) H = I2Rt
6) An incandescent lamp labelled as 10 W,320 V.
a) What do you understand from this?
b) Calculate the energy produced in joules by an incandescent lamp of power 100 W in 10 seconds.
Answer :
a) In 320 V potential difference, the power of the lamp is 100 W.
b) Energy, H = I2Rt= Pt
P = 100 W
H = 100 W x 10 = 1000 J
7) 1000 J heat is generated in a copper wire, when current flow it through one second.
a) If current is doubled, what happens to the heat generated?
b) If current is halved, what happens to the heat generated?
Answer : a) H =
If I is doubled
H = (2I)2 Rt = 4I2Rt
Heat generated becomes 4 times = 4000 J
b) Heat developed reduces to 1/4 th = 250 J
8) So many appliances are there to obtain the fiqhting effect of electric current.
a. Which is the best among them?
b. What are its advantages?
Answer :
a. LED lamps
b. i. As there is no filament, there is no loss of energy in the form of heat.
ii. Since there is no mercury in it, it is not harmful to environment.
iii. It requires only a small quantity of power.
iv. High longevity
9) If 1 A current flows through a conductor of resistance 100 Ω in 1 miniute. What is the heat generated in the conductor.?
Answer : a) H =
H = I2 x 100 x 60 = 6000 J
10). Safety fuse is a device that works on the heating effect of electric current.
a) Which is the main part of a safety fuse.
b) What Is the main advantage of a fuse wire?
Answer : a - Fuse wire
b - Low melting point
11) We are familiar with yellow light and white light in discharge lamps.
a) For this which gases are filled in the discharge lamps.
b) From various lamps, which of them are determental to environment due to the presence of mercury.
Answer : a) For getting white light - mercury
For getting yellow light - Sodium vapour
b) Fluorescent lamps
12) A potential difference of 230 V is applied across a conductor of resistance 200 ft.
a. What is the amount of electric charge passing through this conductor within 5 seconds?
b. Find out the amount of work done to pass the electric charge through this conductor.
Answer :
a. Q = It= (V/R)*t = (230/200) * 5 C = 5.75 C
b. W = VQ = 230 * 5.75 J = 1322.5 J
13) Correct the mistakes, if any, in the underlined parts of the statements given below:
a. The filament in an incandescent lamp has low melting point and high reslstivity.
b. Discharge lamp emits light from the electrodes fitted at its ends.
Answer : (a) High melting point and high resistivity.
b) The gas filled in it.
14) Electrical energy is transformed into mechanical energy in an electric fan.
a. What is meant by the effect of electric current?
b. Write the name of an appliance in which the chemical effect of electric current is used?
Answer : a. The useful form of into which a device converts electrical energy, is considered as the effect of electric current on that device.
b. Storage battery.
15) Depict 2 circuits that able to construct 5ft resistance by connecting four resistors of 5 properly.
Answer :
16)
a).
What is the power of the battery in the circuit?
b).
How can the ammeter readinq be doubled without makinq any chanqe in the battery
or rheostat?
Answer :
a. P = V2/R
P
= (4 x 4 ) /12 = 1.33 W
b.
Connect a resistor of 12Q parallel to the 12Q resistor in the circuit.
Effective resistance is halved and intensity of current is doubled.
17).
a)
What is the wattage of the bulbs In the circuit.
b)
Calculate the amperage of the fuse wire in the circuit.
Answer : a) Wattage = 60 W + 100 W = 160 W
Amperage
= (Wattage/Voltage) = 160/230
= 0.69 A = 1A
Other important links
Class10-Physics-Ch1-Effects
of Electric current - Main reference
Class10-Physics-Ch1-Effects
of Electric current - 1 mark expected question - page-1
Class10-Physics-Ch1-Effects
of Electric current - 1 mark expected question - page-2
Class10-Physics-Ch1-Effects
of Electric current - 2 mark expected questions
Class10-Physics-Ch1-Effects
of Electric current - 3 mark expected questions