Class 10 Science Chapter 11 Electricity Activities

Activity 11.1: Studying Ohm’s Law (V-I Relationship)

Determining the Linear Relationship Between Potential Difference and Current

Experimental Procedure

• Set up a circuit consisting of a nichrome wire \(XY\) of length \(0.5\text{ m}\), an ammeter, a voltmeter, and four cells of \(1.5\text{ V}\) each.
• Use only one cell initially, note the current (\(I\)) from the ammeter and potential difference (\(V\)) from the voltmeter across wire \(XY\).
• Repeat the observations sequentially using two, three, and four cells in the circuit.
• Calculate the ratio of \(V\) to \(I\) for each pair and plot a graph between \(V\) and \(I\).

Observations & Optical Mechanism

• Constant Ratio: Approximately the same value for \(V/I\) is obtained in each case.
• Linear Graph: The \(V-I\) graph is a straight line passing through the origin, showing that potential difference increases linearly with current.
• Ohm’s Law: Confirms that \(V \propto I\) at a constant temperature, where the slope of the line represents the resistance \(R\).

Q1: What is nichrome, and why is it used in this activity?

Answer: Nichrome is an alloy of nickel, chromium, manganese, and iron metals chosen for its stable and measurable resistance.

Q2: What does the straight-line graph passing through the origin signify for a nichrome wire?

Answer: It signifies that the potential difference across the wire is directly proportional to the current flowing through it.

Q3 (Application Based): State Ohm’s mathematical expression and define the unit of resistance.

Answer:

\(V = IR\)

. One ohm (\(1\ \Omega\)) is resistance when a potential difference of \(1\text{ V}\) drives a current of \(1\text{ A}\).

Q4 (Additional Practice): What happens to the current through a resistor if its resistance is doubled while voltage stays constant?

Answer: The current gets halved according to \(I = V/R\).

Q5 (Advanced Mastery): Why must temperature remain constant during verification of Ohm’s law?

Answer: Because the resistance of metallic conductors varies with temperature, which would alter the \(V/I\) ratio.

Activity 11.2: Comparing Resistance of Different Components

Observing Ammeter Variations with Different Gap Components

Experimental Procedure

• Set up a circuit with four dry cells (\(1.5\text{ V}\) each) in series, an ammeter, and a gap \(XY\).
• Connect a nichrome wire in gap \(XY\), plug the key, and note the ammeter reading.
• Replace the nichrome wire with a torch bulb and measure the current.
• Repeat the step using a \(10\text{ W}\) bulb or other material components in gap \(XY\).

Observations & Optical Mechanism

• Current Variation: Ammeter readings differ for different components connected in gap \(XY\).
• Path Resistance: Certain components offer easy paths (good conductors) while others strongly restrain electron motion (resistors or poor conductors).

Q1: Why are ammeter readings different for different components connected in gap \(XY\)?

Answer: Because different materials and components offer varying degrees of resistance to the flow of electrons.

Q2 (Application Based): What microscopic factor causes resistance inside a metallic conductor?

Answer: The motion of electrons is restrained and retarded by the attractive forces of the fixed metal atoms among which they move.

Q3 (Additional Practice): Distinguish between a good conductor, a resistor, and an insulator based on resistance.

Answer: Good conductors offer low resistance, resistors have appreciable resistance, and insulators offer exceptionally high resistance.

Activity 11.3: Factors Affecting Conductor Resistance

Studying the Effect of Length, Area, and Material on Resistance

Experimental Procedure

• Set up a circuit with a cell, ammeter, plug key, and a nichrome wire of length \(l\).
• Plug the key and note the current reading.
• Replace with a nichrome wire of the same thickness but double the length (\(2l\)) and record current.
• Replace with a thicker nichrome wire (larger cross-sectional area \(A\)) and a copper wire of identical dimensions to observe current differences.

Observations & Optical Mechanism

• Length Dependency: Doubling the length halves the ammeter reading, indicating resistance doubles (\(R \propto l\)).
• Area Dependency: Increasing cross-sectional area increases current, showing resistance decreases (\(R \propto 1/A\)).
• Material Dependency: Resistivity (\(\rho\)) varies depending on the nature of the substance.

Q1: State the formula combining length and cross-sectional area to define electrical resistivity.

Answer:

\(R = \rho \frac{l}{A}\)

Q2 (Application Based): What is the SI unit of electrical resistivity, and why are alloys commonly used in heating devices?

Answer: The unit is \(\Omega\text{ m}\). Alloys are used because they have higher resistivity than pure metals and do not oxidize readily at high temperatures.

Q3 (Additional Practice): Calculate the resistance change if a wire’s length is doubled and its cross-sectional area is halved.

Answer: Resistance becomes four times the original value since \(R \propto l/A\).

Activity 11.4 & 11.5: Resistors in Series Combination

Investigating Current and Potential Difference in Series Circuits

Experimental Procedure

• Join three resistors of different values (e.g., \(1\ \Omega, 2\ \Omega, 3\ \Omega\)) in series with a \(6\text{ V}\) battery and an ammeter.
• Move the ammeter to different positions between resistors and note readings.
• Insert a voltmeter across the entire combination (\(X-Y\)) and then across individual resistors (\(X-P\), etc.) to record potential differences.

Observations & Optical Mechanism

• Constant Current: Ammeter shows identical current regardless of its position in a series circuit.
• Additive Potential: Total voltage equals the sum of individual voltage drops (\(V = V_1 + V_2 + V_3\)).
• Equivalent Resistance: \(R_s = R_1 + R_2 + R_3\).

Q1: Why is a series arrangement impracticable for domestic electrical circuits?

Answer: If one component fails, the circuit breaks and all appliances stop working; furthermore, appliances requiring different currents cannot operate together.

Q2 (Application Based): Three resistors of \(2\ \Omega\), \(3\ \Omega\), and \(6\ \Omega\) are connected. How to get total resistance of \(9\ \Omega\)?

Answer: Connect all three in series so that \(R_s = 2 + 3 + 6 = 9\ \Omega\).

Activity 11.6: Resistors in Parallel Combination

Investigating Current Splitting and Equivalent Resistance in Parallel

Experimental Procedure

• Connect three resistors in parallel between points \(X\) and \(Y\) with a battery, ammeter, and plug key.
• Measure the total circuit current (\(I\)) and the potential difference (\(V\)) across the combination.
• Insert the ammeter in series with each individual branch resistor (\(R_1, R_2, R_3\)) to measure branch currents (\(I_1, I_2, I_3\)).

Observations & Optical Mechanism

• Constant Voltage: Potential difference across each parallel branch is identical to the source voltage.
• Additive Current: Total current equals the sum of branch currents (\(I = I_1 + I_2 + I_3\)).
• Parallel Reciprocal Rule: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\).

Q1: What are the main advantages of connecting electrical devices in parallel?

Answer: Each appliance gets the same voltage, operates independently, and failure of one does not affect the others.

Q2 (Application Based): How can resistors of \(2\ \Omega\), \(3\ \Omega\), and \(6\ \Omega\) be combined to yield an equivalent resistance of \(1\ \Omega\)?

Answer: By connecting all three in parallel: \(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1\ \Omega^{-1}\), giving \(R_p = 1\ \Omega\).

Q3 (Advanced Mastery): What happens to total circuit resistance when more resistors are added in parallel?

Answer: The total equivalent resistance decreases, which allows higher total current draw from the source.

Core Concepts: Heating Effect of Current & Electric Power

Joule’s Law of Heating, Fuses, and Power Formulas

Theoretical Principles & Formulas

• Joule’s Law of Heating: Heat produced in a resistor is given by
  
  \(H = I^2Rt\)
• Electric Power: The rate of consumption of electrical energy, expressed as
  
  \(P = VI = I^2R = \frac{V^2}{R}\)
• Commercial Unit: Electrical energy is measured in kilowatt-hour (\(\text{kWh}\)), where \(1\text{ kWh} = 3.6 \times 10^6\text{ J}\).

Practical Applications

• Heating Appliances: Devices like irons, toasters, and heaters use high-resistivity alloy coils.
• Bulb Filaments: Made of tungsten (high melting point \(3380^\circ\text{C}\)) enclosed in inactive nitrogen/argon gas.
• Electric Fuse: Safety wire of low melting point connected in series that melts and breaks the circuit during excessive current surges.

Q1: Why does the cord of an electric heater not glow while the heating element does?

Answer: The heating element has a very high resistance causing extreme heat and glow, whereas the connecting cord has extremely low resistance, generating negligible heat.

Q2: State the commercial unit of electrical energy and convert it into joules.

Answer: Kilowatt-hour (\(\text{kWh}\)), equivalent to \(3.6 \times 10^6\text{ J}\).

Q3 (Application Based): Why is tungsten exclusively used for electric lamp filaments?

Answer: Because it has a very high melting point (\(3380^\circ\text{C}\)) and retains heat to emit bright light without melting.

Q4 (Additional Practice): How does an electric fuse protect domestic appliances?

Answer: When current exceeds a safe rating, the fuse wire’s temperature rises, causing it to melt and disconnect the circuit safely.

Q5 (Advanced Mastery): An electric refrigerator rated \(400\text{ W}\) operates \(8\text{ hours/day}\) for \(30\text{ days}\). Calculate the total energy consumed in \(\text{kWh}\).

Answer: \(\text{Energy} = 0.4\text{ kW} \times 8\text{ h/day} \times 30\text{ days} = 96\text{ kWh}\).