PSEB 10th Science – Chapter 11 Electricity – Super Easy Language

Electricity – Textbook Questions

1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is:
(d) 25.

• Step 1: When the wire of resistance R is cut into five equal parts, the resistance of each new part becomes R/5.

• Step 2: These five parts are connected in parallel. The formula for equivalent resistance (R’) in parallel is:
  1/R’ = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)
  1/R’ = 5 / (R/5)
  1/R’ = 25 / R

• Step 3: Rearranging the formula to find R’, we get:
  R’ = R / 25

• Step 4: Now, find the ratio R/R’:
  R / R’ = R / (R/25) = R * (25/R) = 25.

2. Which of the following terms does not represent electrical power in the circuit?
(b) IR²
The correct formulas for electrical power are P = VI, P = I²R, and P = V²/R. The term IR² is not a valid formula for power.

3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
(d) 25 W.

• Step 1: First, calculate the resistance (R) of the bulb using its rating. The resistance of the bulb does not change.
  Power (P) = V²/R
  R = V²/P = (220 V)² / 100 W = 48400 / 100 = 484 Ω.

• Step 2: Now, calculate the new power (P’) consumed when the bulb is operated at the new voltage (V’ = 110 V).
  P’ = (V’)²/R = (110 V)² / 484 Ω = 12100 / 484 = 25 W.

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. The ratio of heat produced in series and parallel combinations would be:
(c) 1 : 4

• Step 1: Let the resistance of each wire be R.

• Step 2: When connected in series, the total resistance is R_series = R + R = 2R.

• Step 3: When connected in parallel, the total resistance is 1/R_parallel = 1/R + 1/R = 2/R, so R_parallel = R/2.

• Step 4: The formula for heat produced is H = (V²/R) × t. We assume the voltage (V) and time (t) are the same in both cases.
  Heat in series (H_s) = (V² / 2R) × t
  Heat in parallel (H_p) = (V² / (R/2)) × t = (2V² / R) × t

• Step 5: Find the ratio H_s / H_p:
  Ratio = [(V² / 2R) × t] / [(2V² / R) × t]
  Ratio = (V²/2R) × (R/2V²) = 1/4.
  So the ratio is 1:4.

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
A voltmeter is always connected in parallel across the two points where the potential difference is to be measured.

6. A copper wire has a diameter of 0.5 mm and a resistivity of 1.6 × 10⁻⁸ Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Part 1: Finding the length of the wire.

• Given:

  • Resistance (R) = 10 Ω

  • Resistivity (ρ) = 1.6 × 10⁻⁸ Ωm

  • Diameter (d) = 0.5 mm = 0.5 × 10⁻³ m

• Step 1: Calculate the radius (r) of the wire.
  r = d/2 = (0.5 × 10⁻³) / 2 = 0.25 × 10⁻³ m.

• Step 2: Calculate the cross-sectional area (A) of the wire.
  A = πr² = 3.14 × (0.25 × 10⁻³ m)² = 3.14 × 0.0625 × 10⁻⁶ m² = 0.19625 × 10⁻⁶ m².

• Step 3: Use the resistance formula R = ρL/A to find the length (L).
  L = RA / ρ = (10 Ω × 0.19625 × 10⁻⁶ m²) / (1.6 × 10⁻⁸ Ωm)
  L = 122.65 m.
  The length of the wire will be 122.7 m.

Part 2: Change in resistance if the diameter is doubled.

• Resistance (R) is inversely proportional to the cross-sectional area (A), and A is proportional to the square of the diameter (d²). So, R is inversely proportional to d².

• If the diameter is doubled (d’ = 2d), the new resistance (R’) will be:
  R’ ∝ 1 / (2d)² = 1 / 4d²
  This means R’ = R/4.
  If the diameter is doubled, the resistance becomes one-fourth of its original value.

7. The values of current I flowing in a given resistor for corresponding values of potential difference V are given. Plot a graph between V and I and calculate the resistance of that resistor.

Here is how you can plot the graph and calculate the resistance from the given data.

1. Plotting the Graph (V vs. I)

To plot the graph, you would follow these steps:

• Draw the Axes: Draw two perpendicular lines on a graph paper. The horizontal line is the x-axis, and the vertical line is the y-axis.

• Label the Axes: Label the x-axis as “Current (I) in amperes” and the y-axis as “Potential Difference (V) in volts”.

• Choose a Scale: Select a suitable scale for both axes so that all the data points can be plotted.

  • For the x-axis (Current), you could use a scale like 1 cm = 0.5 A.

  • For the y-axis (Voltage), you could use a scale like 1 cm = 2 V.

• Plot the Points: Mark the points on the graph corresponding to the pairs of V and I values:

  • (0.5 A, 1.6 V)

  • (1.0 A, 3.4 V)

  • (2.0 A, 6.7 V)

  • (3.0 A, 10.2 V)

  • (4.0 A, 13.2 V)

• Draw the Line: Draw a straight line that passes through the origin (0,0) and connects as many of the plotted points as possible. This line is the V-I graph. The graph is a straight line, which shows that the resistor obeys Ohm’s Law.

2. Calculating the Resistance

According to Ohm’s Law, resistance (R) is given by the formula R = V/I. The resistance is also the slope of the V-I graph. We can find the resistance by calculating the average value from the given data.

• For the first reading: R₁ = V/I = 1.6 V / 0.5 A = 3.2 Ω

• For the second reading: R₂ = V/I = 3.4 V / 1.0 A = 3.4 Ω

• For the third reading: R₃ = V/I = 6.7 V / 2.0 A = 3.35 Ω

• For the fourth reading: R₄ = V/I = 10.2 V / 3.0 A = 3.4 Ω

• For the fifth reading: R₅ = V/I = 13.2 V / 4.0 A = 3.3 Ω

To find the most accurate value for the resistance, we can take the average of these calculations:

Average Resistance (R) = (R₁ + R₂ + R₃ + R₄ + R₅) / 5
R = (3.2 + 3.4 + 3.35 + 3.4 + 3.3) / 5
R = 16.65 / 5
R = 3.33 Ω

Conclusion:
The resistance of the resistor is 3.33 Ω.

8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

• Given:

  • Voltage (V) = 12 V

  • Current (I) = 2.5 mA = 2.5 × 10⁻³ A

• Solution: Use Ohm’s Law (R = V/I).
  R = 12 V / (2.5 × 10⁻³ A)
  R = 4.8 × 10³ Ω = 4800 Ω.
  The resistance of the resistor is 4800 Ω or 4.8 kΩ.

9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω. How much current will flow through the 12 Ω resistor?

• Step 1: In a series circuit, the total resistance (R_total) is the sum of all individual resistances.
  R_total = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω.

• Step 2: Calculate the total current (I) flowing in the circuit using Ohm’s Law (I = V/R).
  I = 9 V / 13.4 Ω ≈ 0.67 A.

• Step 3: In a series circuit, the current is the same through every component.
  Therefore, the current flowing through the 12 Ω resistor is 0.67 A.

10. How many 176 Ω resistors in parallel are required to carry 5 A on a 220 V line?

• Step 1: First, find the total equivalent resistance (R_eq) the circuit must have using Ohm’s Law.
  R_eq = V / I = 220 V / 5 A = 44 Ω.

• Step 2: Let ‘n’ be the number of 176 Ω resistors needed. For identical resistors in parallel, the formula is R_eq = R / n.
  44 Ω = 176 Ω / n
  n = 176 / 44 = 4.
  4 resistors are required.

11. How will you connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω; (ii) 4 Ω?

• (i) To get 9 Ω: Connect two resistors in parallel and then connect the third resistor in series with this parallel combination.

  • Resistance of two in parallel = (6 × 6) / (6 + 6) = 36 / 12 = 3 Ω.

  • Total resistance = 3 Ω + 6 Ω = 9 Ω.

• (ii) To get 4 Ω: Connect two resistors in series and then connect the third resistor in parallel with this series combination.

  • Resistance of two in series = 6 Ω + 6 Ω = 12 Ω.

  • Total resistance = (12 × 6) / (12 + 6) = 72 / 18 = 4 Ω.

12. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of a 220 V line if the maximum allowable current is 5 A?

• Step 1: Find the current drawn by a single bulb.
  P = VI => I = P / V
  I_bulb = 10 W / 220 V = 1/22 A.

• Step 2: Let ‘n’ be the number of lamps. In a parallel circuit, the total current is the sum of the currents in each branch.
  I_total = n × I_bulb

• Step 3: The total current cannot exceed the maximum allowable current (5 A).
  n × (1/22 A) ≤ 5 A
  n ≤ 5 × 22
  n ≤ 110.
  A maximum of 110 lamps can be connected in parallel.

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

• Case 1: Used separately:

  • R = 24 Ω.

  • I = V/R = 220 V / 24 Ω = 9.17 A.

• Case 2: Used in series:

  • R_total = 24 Ω + 24 Ω = 48 Ω.

  • I = V/R_total = 220 V / 48 Ω = 4.58 A.

• Case 3: Used in parallel:

  • R_total = (24 × 24) / (24 + 24) = 576 / 48 = 12 Ω.

  • I = V/R_total = 220 V / 12 Ω = 18.33 A.

14. Compare the power used in the 2 Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors.

• Step 1: Find the total resistance: R_total = 1 Ω + 2 Ω = 3 Ω.

• Step 2: Find the total current (which is the same through both resistors): I = V/R_total = 6 V / 3 Ω = 2 A.

• Step 3: Calculate the power in the 2 Ω resistor: P = I²R = (2 A)² × 2 Ω = 8 W.

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

• Step 1: In a parallel circuit, the voltage across each resistor is the same as the battery voltage, which is 4 V.

• Step 2: Calculate the power in the 2 Ω resistor: P = V²/R = (4 V)² / 2 Ω = 16 / 2 = 8 W.

• Comparison: The power used in the 2 Ω resistor is the same in both circuits (8 W).

15. Two lamps, one rated 100 W at 220 V and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

• Step 1: In a parallel connection, the total power is the sum of the individual powers.
  P_total = P₁ + P₂ = 100 W + 60 W = 160 W.

• Step 2: Use the power formula P = VI to find the total current (I).
  I = P_total / V = 160 W / 220 V = 16/22 = 8/11 A.
  The total current drawn from the line is approximately 0.73 A.

16. Which uses more energy, a 250 W TV set for 1 hour or a 1200 W toaster for 10 minutes?

• Step 1: Calculate the energy used by the TV set. (Energy = Power × time).

  • Power = 250 W

  • Time = 1 hour = 3600 seconds

  • Energy_TV = 250 J/s × 3600 s = 900,000 J.

• Step 2: Calculate the energy used by the toaster.

  • Power = 1200 W

  • Time = 10 minutes = 10 × 60 = 600 seconds

  • Energy_toaster = 1200 J/s × 600 s = 720,000 J.

• Comparison: 900,000 J > 720,000 J.
  The 250 W TV set used for 1 hour consumes more energy.

17. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

• Analysis: The “rate at which heat is developed” is another term for electrical power. The time (2 hours) is not needed for this calculation.

• Given: R = 8 Ω, I = 15 A.

• Formula: Power (P) = I²R

• Solution: P = (15 A)² × 8 Ω = 225 × 8 = 1800 W.
  Heat is developed at a rate of 1800 Joules per second (or 1800 W).

18. Explain the following:
(a) Why is tungsten used almost exclusively for the filament of electric lamps?
Tungsten is used because it has a very high melting point (3380°C), so it can get white-hot without melting. It also has high resistivity, which helps it to heat up effectively.

(b) Why are the conductors of electric heating devices, such as bread toasters and electric irons, made of an alloy rather than a pure metal?
Alloys (like nichrome) are used because they have a much higher resistivity than pure metals, causing them to generate more heat. They also do not oxidize or burn easily at high temperatures.

(c) Why is a series arrangement not used for domestic circuits?
In a series circuit: 1) If one appliance stops working, the entire circuit is broken, and all other appliances turn off. 2) Each appliance does not get the same voltage; the voltage is divided among them, so they would not work properly.

(d) How does the resistance of a wire vary with its area of cross-section?
The resistance of a wire is inversely proportional to its area of cross-section (R ∝ 1/A). This means a thicker wire (larger area) has less resistance, and a thinner wire (smaller area) has more resistance.

(e) Why are copper and aluminum usually employed for electricity transmission?
Copper and aluminum are used because they have very low resistivity, meaning they are excellent conductors of electricity and lose very little energy as heat. They are also relatively cheap and ductile (can be drawn into wires).

InText Questions

1. What is an electric circuit?
An electric circuit is a closed, continuous path through which an electric current can flow.

2. Define the unit of electric current.
The unit of electric current is the Ampere (A). One Ampere is defined as the flow of one Coulomb of charge through a surface in one second.

3. Calculate the number of electrons that constitute 1 coulomb of charge.

• Given:

  • Total Charge (Q) = 1 C

  • Charge of one electron (e) = 1.6 × 10⁻¹⁹ C

• Solution:

  • Number of electrons (n) = Q / e

  • n = 1 C / (1.6 × 10⁻¹⁹ C) = 6.25 × 10¹⁸.
    One coulomb of charge is constituted by 6.25 × 10¹⁸ electrons.

4. Name a device that helps to maintain a potential difference across a conductor.
A cell or a battery.

5. What is meant by saying that the potential difference between two points is 1 V?
It means that 1 Joule of work is done to move a charge of 1 Coulomb from one point to the other.

6. How much energy is given to each coulomb of charge passing through a 6 V battery?

• Formula: Energy (Work) = Potential Difference × Charge (W = VQ)

• Solution:

  • W = 6 V × 1 C = 6 J.
    6 Joules of energy is given to each coulomb of charge.

7. On what factors does the resistance of a conductor depend?
The resistance of a conductor depends on four factors:

1. Its length (L)

2. Its cross-sectional area (A)

3. The nature of its material (resistivity ρ)

4. Its temperature.

8. Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?
Current will flow more easily through a thick wire. This is because a thick wire has a larger cross-sectional area, which gives it a lower resistance, allowing more current to flow for the same voltage.

9. Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half its former value. What change will occur in the current through it?
According to Ohm’s Law (I = V/R), current is directly proportional to the potential difference. If the potential difference is halved, the current will also decrease to half its former value.

10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
This is because alloys like nichrome have a very high resistivity, which causes them to heat up significantly. They also have a high melting point and do not oxidize (burn) easily at high temperatures.

11. (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
(a) Iron is a better conductor than mercury because it has a lower electrical resistivity.
(b) Silver is the best conductor of electricity.

12. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

13. Redraw the circuit in the above question, putting an ammeter to measure the current and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the reading in the ammeter and the voltmeter?

• Step 1: Calculate total voltage and resistance.

  • V_total = 3 cells × 2 V/cell = 6 V.

  • R_total = 5 Ω + 8 Ω + 12 Ω = 25 Ω.

• Step 2: Calculate the ammeter reading (total current).

  • I = V_total / R_total = 6 V / 25 Ω = 0.24 A.
    The ammeter reading would be 0.24 A.

• Step 3: Calculate the voltmeter reading (voltage across 12 Ω resistor).

  • V₁₂ = I × R₁₂ = 0.24 A × 12 Ω = 2.88 V.
    The voltmeter reading would be 2.88 V.

14. Judge the equivalent resistance when the following are connected in parallel: (a) 1 Ω and 10⁶ Ω; (b) 1 Ω, 10³ Ω, and 10⁶ Ω.
A key rule for parallel resistors is that the equivalent resistance is always smaller than the smallest individual resistance.
(a) For 1 Ω and 10⁶ Ω in parallel, the equivalent resistance will be slightly less than 1 Ω.
(b) For 1 Ω, 10³ Ω, and 10⁶ Ω in parallel, the equivalent resistance will also be slightly less than 1 Ω.

15. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

• Step 1: Find the total equivalent resistance (R_eq) of the three appliances in parallel.

  • 1/R_eq = 1/100 + 1/50 + 1/500

  • 1/R_eq = (5 + 10 + 1) / 500 = 16/500

  • R_eq = 500 / 16 = 31.25 Ω.

• Step 2: The electric iron takes the same current, so it must have the same equivalent resistance.
  The resistance of the electric iron is 31.25 Ω.

• Step 3: Calculate the total current (I) flowing through this equivalent resistance.

  • I = V / R_eq = 220 V / 31.25 Ω = 7.04 A.
    The current through the iron is 7.04 A.

16. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

1. Independent Operation: Each device can be switched on or off independently without affecting the others.

2. Proper Voltage: Each device receives the full voltage from the source and operates correctly.

3. No Total Failure: If one device breaks, the others continue to work because the circuit is not broken.

17. How can three resistors of resistance 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω; (b) 1 Ω?
(a) To get 4 Ω: Connect the 3 Ω and 6 Ω resistors in parallel, and then connect the 2 Ω resistor in series with them.
* R_parallel = (3 × 6) / (3 + 6) = 18 / 9 = 2 Ω.
* R_total = 2 Ω + 2 Ω = 4 Ω.
(b) To get 1 Ω: Connect all three resistors in parallel.
* 1/R_total = 1/2 + 1/3 + 1/6 = (3 + 2 + 1) / 6 = 6/6 = 1.
* R_total = 1 Ω.

18. What is the (a) highest; (b) lowest total resistance that can be secured by a combination of four coils of resistance 4 Ω, 8 Ω, 12 Ω, and 24 Ω?
(a) Highest Resistance: Connect all four resistors in series.
* R_highest = 4 + 8 + 12 + 24 = 48 Ω.
(b) Lowest Resistance: Connect all four resistors in parallel.
* 1/R_lowest = 1/4 + 1/8 + 1/12 + 1/24 = (6 + 3 + 2 + 1) / 24 = 12/24 = 1/2.
* R_lowest = 2 Ω.

19. Why does the cord of an electric heater not glow while the heating element does?
The cord and the heating element are connected in series, so the same current flows through both. Heat produced is given by H = I²Rt. Since the current (I) is the same, the heat produced depends directly on the resistance (R). The heating element is made of a high-resistance alloy (like nichrome), so it gets very hot and glows. The cord is made of a low-resistance material (like copper), so it produces very little heat.

20. Compute the heat generated while transferring 96,000 C of charge in one hour through a potential difference of 50 V.

• Formula: Heat generated (H) = Work done (W) = V × Q.

• Solution:

  • H = 50 V × 96,000 C

  • H = 4,800,000 J or 4.8 × 10⁶ J.
    (The time of one hour is extra information not needed for this calculation).

21. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

• Formula: Heat (H) = I²Rt

• Solution:

  • H = (5 A)² × 20 Ω × 30 s

  • H = 25 × 20 × 30 = 15,000 J.
    The heat developed is 15,000 Joules.

22. What determines the rate at which energy is delivered by a current?
The rate at which energy is delivered by a current is called electric power.

23. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

• Part 1: Power of the motor.

  • Power (P) = V × I

  • P = 220 V × 5 A = 1100 W.
    The power of the motor is 1100 W or 1.1 kW.

• Part 2: Energy consumed in 2 hours.

  • Energy (E) = Power × time

  • E = 1.1 kW × 2 h = 2.2 kWh (kilowatt-hours).
    The energy consumed is 2.2 kWh.

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