4.3.2 Series and parallel circuits (3)
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1.
Question 1
A circuit consists of a 9V battery connected to two resistors in parallel. One resistor has a resistance of 2Ω and the other has a resistance of 4Ω. Calculate the total current flowing from the battery.
Concept Used: The sum of the currents entering a junction in a parallel circuit is equal to the sum of the currents that leave the junction. The voltage across parallel resistors is the same. Therefore, we can calculate the current through each resistor and then sum them.
Solution:
- Calculate the current through the 2Ω resistor: Using Ohm's Law (V = IR), the current (I) through the 2Ω resistor is: I = V/R = 9V / 2Ω = 4.5A.
- Calculate the current through the 4Ω resistor: Similarly, the current through the 4Ω resistor is: I = V/R = 9V / 4Ω = 2.25A.
- Calculate the total current: The total current is the sum of the currents through each resistor: 4.5A + 2.25A = 6.75A.
Answer: The total current flowing from the battery is 6.75A.
2.
Describe a practical experiment you could carry out to verify the junction rule. Include a list of materials, the procedure, and how you would record your results. Explain what your results would demonstrate.
Materials:
- Battery (e.g., 9V)
- Battery holder
- Connecting wires
- A switch
- A variable resistor (rheostat)
- An ammeter
- A circuit diagram (see below)
Procedure:
- Construct a circuit according to the diagram below. The rheostat will allow you to vary the resistance in the circuit.
- Connect the ammeter in series with one of the wires in the circuit.
- Set the rheostat to a low resistance.
- Close the switch and record the ammeter reading (current, I).
- Increase the resistance of the rheostat in steps, recording the ammeter reading for each value. Take at least 5 readings.
Circuit Diagram:
| + Battery |
| | |
| --- Rheostat --- Ammeter --- Wire --- Junction --- Wire --- Battery - |
Results Recording: Create a table to record your results:
| Resistance (Ω) | Current (A) |
| 10 | 0.5 |
| 20 | 0.4 |
| 30 | 0.3 |
| 40 | 0.25 |
Demonstration: The results will demonstrate that as the resistance of the circuit increases, the current through the ammeter decreases. However, the current remains constant at the junction. This is because the current entering the junction is the same as the current leaving the junction, as required by the junction rule. The rheostat allows us to control the resistance, but the total current flowing through the junction remains the same.
3.
A student connects three resistors, R, R = 10Ω, and R = 20Ω, in parallel to a 6V battery. The equivalent resistance of the parallel combination is found to be 4Ω. Verify this result and calculate the total current flowing from the battery.
Solution:
- Calculate the equivalent resistance of the parallel combination:
1/Req = 1/R + 1/R + 1/R
1/Req = 1/10Ω + 1/20Ω + 1/20Ω
1/Req = 2/20Ω + 1/20Ω + 1/20Ω = 4/20Ω = 1/5Ω
Req = 5Ω
- The student states the equivalent resistance is 4Ω. This is incorrect. The correct equivalent resistance is 5Ω.
- Calculate the total current (Itotal) from the battery using Ohm's Law:
Itotal = V / Req
Itotal = 6V / 5Ω = 1.2 A
Answers: The equivalent resistance is 5Ω, not 4Ω. Total current = 1.2 A