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Class 10 Physics (India)
Course: class 10 physics (india) > unit 3.
- Electricity class 10 numerical: CBSE board practice
Electricity class 10: CBSE previous question paper problems
Electric current and circuit
| Closed circuits | Open circuits |
|---|---|
| There is a flow of charge. | There is no flow of charge. |
| The switch is ON. | The switch if OFF. |
- I = 10 A ,
- t = 2 mins = 120 s .
- net charge, Q = 1200 C ,
- charge of electron, e = 1.6 × 10 − 19 C ,
- number of electrons, n = ?
- Understand the concept better with this video : Unit of charge (Coulombs)
Electric potential and potential difference
- W = 100 J ,
- Q = 20 C .
- Get more practice with this exercise : Voltage and work
- Understand the concept better with these videos : ( i ) Intro to potential difference (& voltage) , ( ii ) Solved example: Potential difference & work done
Circuit diagram
- Solution A4. [ 1 mark]
- Solution A5. ( i ) Electric cell: [ 0.5 marks] ( ii ) Open plug key: [ 0.5 marks] ( iii ) Wires crossing without connection: [ 0.5 marks] ( iv ) Variable resistor: [ 0.5 marks] ( v ) Battery: [ 0.5 marks] ( vi ) Electric bulb: [ 0.5 marks] ( vii ) Resistance: [ 0.5 marks]
Ohm's law
- Ohm's law: The electric current flowing through a conducting wire, I , is directly proportional to the potential difference across its ends, V , provided its temperature stays unchanged. Mathematically, V ∝ I .
- The graph obtained by plotting the potential difference against the current flowing in a conductor is a straight line .
- From Ohm's law, V ∝ I , or V = I R , where R is the resistance. So, R = V I , which represents the slope of the V − I graph.
- Understand the concept better with these videos : ( i ) Introduction to circuits and Ohm's law , ( ii ) Solved example: Ohms law , ( iii ) Ohm's law graph (verifying Ohm's law) , ( iv ) Solved example: (Ohm's law graph) .
Factor on which the resistance of a conductor depends
- Solution A7. ( a ) The resistance of a conductor: ( i ) is directly proportional to the length of the conductor, l . R ∝ l ( ii ) is inversely proportional to the cross-sectional area of the conductor, A . R ∝ 1 A ( iii ) depends on the material of the conductor. [ 1 mark] ( b ) Metals have a much lower resistivity than glass. This makes metals better conductors of electricity. [ 1 mark] ( c ) Alloys have a high resistivity and do not oxidise or burn readily at high temperatures , making them perfect for use in heating devices. [ 1 mark]
- radius of wire, r = 0.01 cm = 10 − 4 m ,
- resistivity, ρ = 50 × 10 − 8 Ω m ,
- resistance, R = 10 Ω .
- length of wire, l = ?
- length of wire, l = 50 cm = 50 × 10 − 2 m ,
- cross-sectional area, A = 0.01 mm 2 = 0.01 × 10 − 6 m 2 ,
- resistivity, ρ = 5 × 10 − 8 Ω m .
- Understand the concept better with this video : Resistivity and conductivity
Resistance of system of resistors
- Resistance, R 1 = 5 Ω ,
- Resistance, R 2 = 10 Ω ,
- e.m.f., V = 6 V .
- R 1 = R 2 = R 3 = 2 Ω ,
- potential difference, V = 5 V ,
- current, I = ?
- Solution A12. Derivation of equivalent parallel resistance: The following diagram shows a parallel combination of the three resistors R 1 , R 2 , and R 3 . Currents I 1 , I 2 , and I 3 flow through the resistors R 1 , R 2 , and R 3 respectively. The total current, I , is the sum of the currents through the three branches: I = I 1 + I 2 + I 3 If the equivalent resistance is R P , we have using Ohm's law, I = V R P The potential difference across each individual resistor is also V . Applying Ohm's law for each resistor, we find, I 1 = V R 1 , I 2 = V R 2 , and I 3 = V R 3 [ 1 mark] Tying it all up, I = I 1 + I 2 + I 3 V R P = V R 1 + V R 2 + V R 3 V × ( 1 R P ) = V × ( 1 R 1 + 1 R 2 + 1 R 3 ) 1 R P = 1 R 1 + 1 R 2 + 1 R 3 ∴ The equivalent resistance, R P , is related to the individual resistors as: 1 R P = 1 R 1 + 1 R 2 + 1 R 3 [ 1 mark]
- Get more practice with these exercises : ( i ) Finding equivalent resistance , ( ii ) Identifying types of resistor combinations , ( iii ) Simplifying resistor networks , ( iv ) Finding currents and voltages (pure circuits) .
- Understand the concept better with these videos : ( i ) Series resistors , ( ii ) Parallel resistors (part 1) , ( iii ) Parallel resistors (part 2) , ( iv ) Parallel resistors (part 3) , ( v ) Example: Analyzing a more complex resistor circuit , ( vi ) Solved example: Finding current & voltage in a circuit .
Heating effect of electric current
- Solution 1 (in words) A13 (in words). The heat produced every second in a fuse wire of resistance, R , and with a current, I , in it, is given as, H = I 2 R . [ 1 mark] The heat produced for I = 5 A melts the wire. It follows from the relation that for the same heat to be produced for I = 10 A , the resistance must be smaller . [ 1 mark] Since, R ∝ 1 cross-sectional area , a smaller resistance has a larger cross-section . A larger cross-section implies a larger radius . So, the new fuse wire has a larger radius. [ 1 mark]
- Solution 2 (calculation) A13 (calculation). For current I 1 = 5 A passing through the older fuse of resistance R 1 , the heat produced every second is, H = I 1 2 R 1 The same heat is desired for I 2 = 10 A . If the new fuse wire has a resistance, R 2 , we have, H = I 1 2 R 1 = I 2 2 R 2 ⇒ R 2 = I 1 2 I 2 2 R 1 = 25 1 A 100 4 A R 1 = 1 4 R 1 [ 1 mark] ∵ R ∝ 1 A ⇒ 1 A 2 = 1 4 × 1 A 1 ⇒ A 2 = 4 A 1 where A is the cross-sectional area. [ 1 mark] Since, A = π r 2 , we get from the result above, π r 2 2 = 4 π r 1 2 ⇒ r 2 = 2 r 1 The radius of the new fuse wire is twice that of the old one. [ 1 mark]
- Solution A14. Derivation: For work done, W , on moving a net charge, Q , the potential difference is defined as, V = W Q ⇒ W = V Q Let t be the time it takes to move the net charge Q . Multiplying and dividing the R.H.S. by t , we get, W = V × Q t × t = V I t where Q t is the current, I , by definition. [ 1 mark] Since this work is converted into heat energy, we can write, W = H = V I t = ( I R ) × I t H = I 2 R t where R is the resistance in the circuit and Ohm's law ( V = I R ) is applied in the second step. [ 1 mark] Second part: We obtained, W = H = V I t . Writing I = V R from Ohm's law, we get, H = V × ( V R ) × t = V 2 R t H ∝ 1 R [ 1 mark] If two equal resistances, R each, are connected in series , the equivalent resistance, R S = R + R = 2 R . When connected in parallel , the equivalent resistance, R P , is found as, 1 R P = 1 R + 1 R R P = R 2 [ 1 mark] Let the heat produced in the series combination be H S , and the parallel combination be H P . We have, H S H P = 1 / R S 1 / R P = R P R S = R 2 × 1 2 R H S = 1 4 H P H P = 4 H S Therefore, the heat produced in the parallel combination is four times that of the series combination. [ 1 mark]
- Get more practice with this exercise : Calculating heat dissipated in circuits .
- Understand the concept better with these videos : ( i ) Heating effect of current , ( ii ) Solved example - Calculating power & heat dissipated .
Electric power
- Solution A15. Since the lamps are connected in parallel to the mains supply, the voltage across each lamp is 220 V . The current drawn by a lamp = power rating voltage applied . Current drawn by the 100 W lamp, I 1 = 100 W 220 V = 100 220 A Current drawn by the 60 W lamp, I 2 = 60 220 A [ 1 mark] Net current drawn from the mains supply, I , is, I = I 1 + I 2 = 100 220 A + 60 220 A = 160 220 A ≈ 0.73 A [ 1 mark]
- Solution A16. ( a ) Power, P = V I , where V is the voltage and I is the current. Power consumed is minimum when the current passing is minimum. So the resistors should be connected in series . [ 1 mark] ( b ) Bulbs: Power of each bulb = 100 W . Total power of three bulbs, P B = 300 W . Energy consumed by three bulbs in a day, E B = 300 W × 5 h = 1500 Wh = 1.5 kWh [ 1 mark] Fans: Power of each fan = 50 W . Total power of two bulbs, P F = 100 W . Energy consumed by two fans in a day, E F = 100 W × 10 h = 1000 Wh = 1 kWh [ 1 mark] Heater: Energy consumed by heater in a day, E H = 1 kW × 1 2 h = 0.5 kWh [ 1 mark] Total energy consumed, E = E B + E F + E H = ( 1.5 + 1 + 0.5 ) kWh = 3 kWh [ 0.5 marks] Energy consumed in a month of 31 days = E × 31 = 93 kWh . Therefore, the cost of energy consumed ₹ ₹ = ₹ 3.60 / kWh × 93 kWh = ₹ 334.80 [ 0.5 marks]
- Get more practice with this exercise : Bulbs connected in series or parallel .
- Understand the concept better with these videos : ( i ) Electric power & energy , ( ii ) Solved example: Power dissipated in bulbs , ( iii ) Solved example - Cost of operation of electrical device .
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Class 10 Science Chapter-wise assignments Pdf
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Class 10 Science chapter-wise printable assignments Pdf download links
| CLASS 10th SCIENCE ASSIGNMENTS PDF DOWNLOAD | DOWNLOAD LINK |
|---|---|
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| Chapter 4 – Carbon and Its Compounds assignment pdf | |
| Chapter 5 – Periodic Classification Of Elements assignment pdf | |
| Chapter 6 – Life Processes assignment pdf | |
| Chapter 7 – Control and Coordination assignment pdf assignment pdf | |
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| Chapter 10 – Light Reflection and Refraction assignment pdf | |
| Chapter 11 – The Human Eye and Colourful World assignment pdf | |
| Chapter 12 – Electricity assignment pdf | |
| Chapter 13 – Magnetic Effects of Electric Current assignment pdf | |
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