Kerala Syllabus SAMAGRA SCERT SAMAGRA Question Pool for Class 10 English Medium Physics Effects of Electric Current
A filament Lamp designed to work at a potential difference 250V has power 100W. What will be the power of this lamp when connected to a 100V supply?
We know ,power P = V^{2}/R
R = V^{2}/P = 250 x 250 /100 = 625 W
When connected to 100 V power supply ,
Power P =
A heating coil of 10000? resistance works in 250V supply.
Nowadays LED Lamps are widely used .
A,B,C,D & E are five 10Ω resistors connected in a circuit as in the given diagram.
1) Effective resistance of A and B, R1 =A+B=10+10=20 ohm
Effective resistance of c and D, R2 = C + D= 10 +10 = 20 ohm
Effective resistance
Resistance of a 20cm long conductor is 20?. The conductor is bent into circular loops and connected in the circuit as in the given diagrams, Calculate the resultant resistance in each case.
Fig (i) Effective resistance = 20 Ohms Fig(ii) two 10Resistances are connected as parallel so Effective resistance, 1/R = 1/R1 + 1/R2 =
Power of a bulb which works in 220V is 100W. When the voltage in the circuit decreases the power becomes 25W, What will be the voltage at that time?
P = V^{2}/R
R = V^{2}/P = 220 x 220 /100 = 484 Ω
Match the following related to LED Lamp
LED Lamps save energy & are ecofriendly. Justify this statement
Calculate the amount of heat energy produced when 1A current flows through a 1? resistance wire for 1 hour.
Given That,
Resistance, R= 1?
Current,I=1 A
Time, t=1 hour=3600
An electrical device of power 440W is connected to 230V power supply. Which among the following is the amperage of fuse to be used in this circuit ?
(a) 0.5A (b) 2A (c) 1.5A (d) 4A
Amperage = wattage/voltage
=440/230
The given experiment is based on heating effect of electric current
Aluminium and Nichrome wires of same length and thickness are used in the given circuit.
Analyse the diagram and answer the following questions.
If a bulb labelled as 100W/230V is connected to 115V power supply, What will be its Power ? (100W,25W, 12.5W, 50W)
P= 25W
Observe the given circuit Diagram. B1 is a torch bulb and B2 is an ordinary incandescent bulb.
When excess electric current flows through the circuit, fuse wire melts & breaks the circuit.
Find the relation in the first then complete the second pair
Heat energy produced in a current carrying conductor is equal to the product of square of current through the conductor, Resistance of the conductor and the time for which the current flows.
Find the relation and complete the following: Electrical energy→ Heat energy → Heating effect → Electric Stove Electrical energy →Chemical energy → Chemical effect → .....................
Storage Battery
A and B are two electrical devices,
Device A | Device B |
230V | 230V |
1000W | 50W |
Find the correct answer from the following: The device works on the heating effect of electric current.
Fuse
240V power supply is maintained in household circuits.
Match suitably:
The cautionary measures are given while the fuse wire is included in the circuit.
In an electric heater 800W , 400V is labelled.
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