4 Most FAQ’s of Electric Current Chapter in Class 10th Physical Science (TS/AP)

8 Marks

LAQ-1 : Explain Kirchhoff’s laws with examples.

Introduction

Kirchhoff’s laws provide a comprehensive description of the relationship between current and voltage in an electrical circuit, which is critical for analyzing and designing complex circuits.

Kirchhoff’s Current Law (KCL) or Junction Law:

1. Statement:
   The algebraic sum of currents entering a junction (or node) in a circuit is always zero. This is based on the principle of conservation of electric charge.
2. Mathematically:
   $$\sum I_{\text{in}} = \sum I_{\text{out}}$$
3. Example:
   Consider a junction where three wires meet. If currents I₁​, I₂​, and I₃​ are entering the junction and currents I₄​ and I₅​ are leaving the junction, according to KCL: I₁ + I₂ + I₃ = I₄ + I₅​

Kirchhoff’s Voltage Law (KVL) or Loop Law:

1. Statement:
   The algebraic sum of all voltages (potential differences) in a closed loop or mesh in a circuit is always zero. This is based on the principle of conservation of energy.
2. Mathematically:
   $$\sum V = 0$$
3. Example:
   Consider a simple circuit loop containing a battery of 9V and two resistors R₁​ and R₂​. If the voltage drop across R_1​ is 3V and across R₂​ is 6V, according to KVL:
   $$V_{\text{battery}} – V_{R1} – V_{R2} = 0$$
   or
   $$9V – 3V – 6V = 0$$

Summary

Kirchhoff’s laws are fundamental for circuit analysis. The Current Law ensures that current is conserved at any junction in a circuit, reflecting the conservation of electric charge. The Voltage Law ensures that energy is conserved in any closed loop, indicating that the energy supplied to the loop is equal to the energy taken from the loop. These laws are applicable to both AC and DC circuits and are foundational for electrical engineering studies.

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LAQ-2 : Mention the required material in the experiment to verify Ohm’s law and write the experimental procedure.

Introduction

This experiment aims to show how the ratio between voltage (V) and current (I) remains constant for certain materials, but not for others.

Materials Needed

1. Ammeter
2. Voltmeter
3. Five 1.5 V dry cells or batteries
4. Conducting wires
5. An iron or manganin spoke (10 cm in length)
6. Switch or key
7. Light Emitting Diode (LED)

Testing with Iron Spoke

1. Set Up: Set up the circuit as guided.
2. Circuit Completion: Complete the circuit using the iron spoke.
3. Recording Readings: Record the current and voltage readings from the ammeter and voltmeter.
4. Adding Batteries: Add two batteries to the circuit and note the new readings.
5. Continuous Addition and Notation: Continue adding more batteries, one by one, noting the readings each time.
6. Ratio Calculation: Calculate the V/I ratio for each set of readings.

Observation with Iron Spoke

For the iron spoke, the V/I ratio remained stable, indicating a direct relationship between voltage and current.

Testing with LED

1. Replace with LED: Replace the iron spoke with the LED in the circuit.
2. Repeat Process: Repeat the same process as before, noting the readings as you add each battery.
3. Ratio Calculation with LED: Calculate the V/I ratio for each set of readings with the LED.

Observation with LED

The V/I ratio did not stay the same with the LED, showing it varies for different materials.

Results

1. With the iron spoke, there was a direct proportion between voltage and current, given that the temperature remained unchanged.
2. For the LED, the V/I ratio wasn’t consistent.

Summary

The relationship between voltage and current is not the same for all materials. While some, like the iron spoke, maintain a constant V/I ratio, others, such as the LED, do not. This suggests that only specific materials have a steady V/I ratio at a consistent temperature.

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LAQ-3 : List out the material required for the experiment “The effect of increasing of cross-section of a conductor upon its resistance” and write the experimental procedure.

Introduction

This experiment aims to demonstrate the relationship between the resistance of a conductor and its cross-sectional area. It is believed that as the cross-sectional area of a conductor increases, its resistance decreases.

Materials Needed

1. Ammeter
2. Key or switch
3. Battery
4. Conducting wire
5. Iron rods (All should be of equal length but different cross-sectional areas)

Experimental Procedure

1. Setting up the Circuit:
   Assemble the circuit as instructed, ensuring one of the iron rods is connected between points P and Q.
2. Measurement:
   Activate the circuit and measure the current flowing through the iron rod using the ammeter. Record this current value.
3. Testing Different Rods:
   Replace the current iron rod with another one, having a different cross-sectional area. Note down the current readings for each rod you test.
4. Observation:
   As you test rods with larger cross-sectional areas, you will notice that more current flows through them. This implies the rod’s resistance is decreasing.

Results and Conclusion

The experiment confirms that the resistance of a conductor decreases as its cross-sectional area increases, given that the temperature and the length of the conductor remain constant. Mathematically, this can be expressed as

$$R \propto \frac{1}{A}$$

Summary

A conductor’s resistance and its cross-sectional area are inversely related. For students studying this topic, it’s crucial to understand this relationship as it plays a fundamental role in electronics and electrical circuits.

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LAQ-4 : 12 V battery is connected in a circuit and to this 4Ω, 12Ω resistors are connected in parallel, 3Ω resistor is connected in series to this agreement. Draw the electric circuit from this information and find the current in the circuit.

Introduction

This exercise seeks to understand how to set up an electrical circuit based on the provided components and their configurations. We’ll also determine the current flowing in the circuit.

Materials and Setup

1. A 12 V battery.
2. Resistors of values: 4Ω, 12Ω, and 3Ω.

Circuit Diagram

1. Begin with the 12 V battery.
2. Connect the 4Ω and 12Ω resistors in parallel to the battery. This means their starting and ending points are directly connected.
3. Next, connect the 3Ω resistor in series to this parallel configuration. This means one end is connected to the endpoint of the parallel resistors and the other end goes back to the battery.

Calculations

1. Calculating Equivalent Resistance for Parallel Configuration:
   • Formula for resistors in parallel: $$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$$
   • Plugging in the values: $$\frac{1}{R_{\text{parallel}}} = \frac{1}{4} + \frac{1}{12}$$
2. Calculating Total Resistance of the Circuit:
   • Since the 3Ω resistor is in series with the parallel combination, their resistances add up.
   • Total Resistance, $$R_{\text{total}} = R_{\text{parallel}} + 3Ω$$
3. Calculating the Current:
   • Using Ohm’s law, $$I = \frac{V}{R_{\text{total}}}$$​

Summary

In a circuit with a 12V battery connected to a 4Ω and 12Ω resistor in parallel, and a 3Ω resistor in series, the total current flowing through the circuit is calculated using the above steps.