Physics : Class X Chapter: ELECTRICITY Assignment 2012-13
ASSIGNMENT -1, (2012-13) By JSUNIL Sub. – Physics, Class - X Chapter -ELECTRICITY
One mark questions 1. Define the term “electric current”.
2. Define the term ‘resistivity’ of a material.
3. How is a Voltmeter connected in the circuit to measure the potential difference between two points?
4. You have two metallic wires of resistances 6 ohm and 3 ohm. How will you connect these wires to get the effective resistance of 2 ohm?
5. If the distance between two electric charges is doubled, how much will the force exerting between them change to? Two marks questions
6. State Ohm’s law. “The resistance of a conductor is 1Ω.” What is meant by this statement?
7. Why are coils of electric toaster made of an alloy rather than a pure metal? 8. Why is... Read more »
5. If the distance between two electric charges is doubled, how much will the force exerting between them change to? Two marks questions
6. State Ohm’s law. “The resistance of a conductor is 1Ω.” What is meant by this statement?
7. Why are coils of electric toaster made of an alloy rather than a pure metal? 8. Why is... Read more »
Problem : 1
ReplyDeleteThree resistances of values 2Ω,3Ω and 5Ω are connected in series across 20 V,D.C supply .Calculate (a) equivalent resistance of the circuit (b) the total current of the circuit (c) the voltage drop across each resistor and (d) the power dissipated in each resistor.
Total resistance R = R1 + R2+ R3.
= 2 +3+5 = 10Ω
Voltage = 20V
Total current I = V = 20 = 2A.
R 10
Voltage drop across 2Ω resistor V1 = I R1
= 2× 2 = 4 volts.
Voltage drop across 3Ω resistor V2 = IR2
= 2 × 3 = 6 volts.
Voltage drop across 5Ω resistor V3 = I R3
= 2 ×5 = 10 volts.
Power dissipated in 2Ω resistor is P1 = I2 R1
= 22 × 2 = 8 watts.
Power dissipated in 3 resistor is P2 = I2 R2.
= 22 × 3 = 12 watts.
Power dissipated in 5 resistor is P3 = I2 R3
= 22 × 5 = 20 watts.
Problem :2
A lamp can work on a 50 volt mains taking 2 amps.What value of the resistance must be connected in series with it so that it can be operated from 200 volt mains giving the same power.
Lamp voltage ,V = 50V
Current ,I = 2 amps.
Resistance of the lamp = V = 50 = 25 Ω
Resistance connected in series with lamp = r.
Supply voltage = 200 volt.
Circuit current I = 2A
Total resistance Rt= V = 200 = 100Ω
I 2
Rt = R + r
100 = 25 + r
r = 75Ω 8