Physics Chapter 1
Effects of Electric current
3 Marks Expected Question
(1) Complete the table
|
R (Q) |
V(V) |
t(s) |
1 (A) |
H |
Change in H |
|
R |
V |
t |
1 |
H |
|
|
2R |
V |
t |
1/2 |
(a) |
(b) |
|
R |
2V |
t |
21 |
4H |
(c) |
|
R |
V |
2t |
1 |
(d) |
(e) |
|
R/2 |
(f) |
2t |
41 |
16H |
16 times |
Answer:
a) H/2
b) made half
c) becomes 4 times
d) 2H
e) becomes doubled
f) 2V
(2) Find the relation.
|
a)lncandescent lamp |
Low power |
|
b) LED |
Carbon rod |
|
c) Fluorescent lamp |
Electrical energy lost as heat |
|
d) Arc lamp |
Heating coil coated with thorium oxide. |
Answer:
|
a)lncandescent lamp |
Electrical energy lost as heat |
|
b) LED |
Low power |
|
c) Fluorescent lamp |
Heating coil coated with thorium oxide. |
|
d) Arc lamp |
Carbon rod |
(3) Features of some electric lamps are given below. Fill the blanks appropriately.
|
Feature |
lamp |
|
The Inconveniece caused by the shadow
is minimised. |
1)___________ |
|
Gives yellow light |
2)___________ |
|
Electrical energy is lost as heat. |
3)___________ |
Answer:
1) Fluorescent lamp
2)
Discharge lamp with sodium vapor
c)
Filament lamp
(4) When a heater with a certain resistance was connected to a 200 V supply, 120000 J heat was generated.
Find Quantity of electric charge passed through the heater ?
Find out the resistance of the heater ?
Answer:
V = 200 V, t = 5 * 60 s = 300 s
H = 120000
H = VIt
H 120000
I=H/Vt =120000/(200x300)
I = 2A
Charge 0 = It = 2 x 300 = 600 C
or we can say
Quantity of heat generated H = VQ
Q = H/V = 120000/200 = 600 C
Answer:
Another Method
To find the quantity of electric charge that passed
through the heater, you can use the formula:
charge = energy / voltage
Substituting the given values, we get:
charge = 120000 J / 200 V
charge = 600 Coulombs
To find the resistance of the heater, you can use the formula:
resistance = voltage / current
Since we know the voltage and the charge, we can use the formula for electric current:
current = charge / time
Substituting this into the formula for resistance, we get:
resistance = voltage / (charge / time)
If we assume that the charge passed through the heater in a time of 1 second, we can solve for the resistance:
resistance = 200 V / (600 C / 1 s)
resistance = 200 V / 600 C/s
resistance = 0.33 ohms
(5) 'Damaged LED bulbs cause the formation of e-wastes'.
What are the damages?
Answer:
1. An LED bulb is a combination of many LEDs
connected in series. The bulb will fail to emit light if any one LED is damaged
or if the contact of any one LED is lost.
2. LED bulb will not emit light if the rectifier or
load resistor or filter capacitor is damaged.
3. Any damage in the bulb circuit.
(6) Resistance of an electric heater working in 230V is 1000 Ohm.
a) Write down the change of energy in electric heater.
b) Calculate the current utilized by the heater in one hour.
a) Electrical energy converted in to heat energy.
Power = Voltage * Current
Plug in the known values of Voltage and Resistance,
Power = 230V * (1000 Ohm)
b) To find the current utilized by the heater in one hour,
we first need to find the current in amperes (A) by using the formula:
Power = Voltage * Current
Current = Power / Voltage
Then, we can use the formula Current x Time = Charge
to find the charge passed in one hour.
Charge = Current x Time, where time is in seconds and
convert the seconds to hours by dividing the seconds to 3,600.
As, Current = Power / Voltage, thus we can directly
calculate the current utilized by the heater in one hour as Current = Power /
Voltage * (3600 sec/1 hour)
(V2t)/R = (230 x 230 x 60 x 60) / 1000
= 190440 J
(7) Find
the relation.
|
a)lncandescent
lamp |
Low
power |
|
b)
LED |
Carbon
rod |
|
c)
Fluorescent lamp |
Electrical
energy lost as heat |
|
d)
Arc lamp 0 |
Heating
coil coated with thorium oxide. |
Answer:
|
a)
Incandescent lamp |
Electrical
energy lost as heat |
|
b)
LED |
Low
power |
|
c)
Fluorescent lamp |
Heating
coil coated with thorium |
|
d)
Arc lamp |
Carbon
rod |
(8) Write the effect of electricity in each of the following.
a) Working of electromagnet
b) Electric bulb
c) Electric heater
Answer:
a) Magnetic effect
b) Lighting effect
c) Heating effect
(9) 'Damaged LED bulbs cause environmental problems'
a. What Is the environmental problem caused by this?
b. How can this be solved?
c. Which parts of an LED bulb become damaged?
Answer:
a. Disposal of e-waste
b. Segregate the plastic, electronic and metal
components and transfer them to their disposal units.
c. Rectifier, Load resistor, Filter capacitor and LED
chip.
(10) Features
of some electric lamps are given below. Fill the blanks appropriately.
|
Feature |
lamp |
|
The
Inconveniece caused by the |
1)....... |
|
Gives
yellow light |
2)........ |
|
Electrical
energy is lost as heat. |
3)........ |
Answer:
1)
Fluorescent lamp
2)
Discharge lamp with sodium vapour
c)
Filament lamp
(11) Safety fuse is a device which is used to protect an
electric circuit.
a) What are the features of a fuse wire ?
b) What are the precautions to be taken when a fuse is
connected in a house hold circuit ?
Answer:
Low melting point, an alloy of tin and load, possess appropriate amperage.
The ends of the fuse wire must be connected firmly at appropriate points.
The fuse wire should not project out of the carrier base.
Fuse wire must be connected in between the terminals of the porcelain bridge.
Don't use thick wire or copper wire as fuse wire.
a. What properties of tungsten make it suitable for
being used as a filament?
b. What method Is used to reduce vaporisation of
tungsten filament?
c. What is the main disadvantage of incandescent
lamps?
Answer:
a. i. High resistivity
ii. High melting point
iii. Ductility
iv. Ability to emit white light in the while hot
condition.
b. Vaporisation can be reduced by filling some inert
gas or nitrogen at low pressure inside the bulbs.
c. A major part of the electrical energy supplied to
an incandescent lamp lost as heat.
(13)
a)What is meant by the amperage of an electric
appliance?
b) Calculate the amperage of an electric iron of power
1000 W
c) What is the relation between the thickness and the
amperage of a conductor.
Answer:
a) Amperage is the ratio of the power of an equipment
to the voltage applied (p/v).
b) Amperage = P/V = 1000 / 230 = 43 A
c) Amperage increases with the thickness of the conductor.
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