Electric Circuits: Series and Parallel

Electric Components and Symbols

Open switch: A switch that does not allow current to flow.
Closed switch: A switch that allows current to flow.
Cell: A single unit providing electrical energy.
Battery: Multiple cells connected together.
Voltmeter: An instrument for measuring voltage (V).
Ammeter: An instrument for measuring current (A).
Lamp: A device that emits light when current flows through it.
Resistor: A component that opposes the flow of current.
Voltmeter in Circuit: Connected in parallel with the component to measure voltage across it.
Ammeter in Circuit: Connected in series with the component to measure current through it.

Measuring Voltage

Series Circuit

Identical Lamps: If three identical lamps are connected in series, the voltage of the battery is split equally between them.
Different Components: If different components (e.g., a lamp and a buzzer) are in series, the voltage across each will be different.

Parallel Circuit

Voltage: The voltage across each branch of a parallel circuit is equal to the voltage of the supply, regardless of the components in each branch.

Measuring Current

Series Circuit

Current: The current is the same at all points in a series circuit.

Parallel Circuit

Current Division: At a branch in a parallel circuit, the current splits.
Equal Lamps: If two identical lamps are in parallel, the current will divide equally between them.
Current Combination: When branches rejoin, the current combines again to equal the original current.

Practice Questions

Question 1: Missing currents in various parallel circuits need to be calculated using the principle that total current entering a junction equals total current leaving the junction.
- Example circuit values: 3.5 A, 5.5 A, 1.0 A, 0.5 A, 3.0 A, 1.5 A
Question 2: Missing currents in various parallel circuits need to be calculated using the principle that total current entering a junction equals total current leaving the junction.
- Example circuit values: 1 A, 2 A, 1.5 A, 0 A, 1.0 A, 0.5 A
Question 3: Zara's Series Circuit
- Given: 9V battery, a lamp (L₁) and a buzzer in series; voltage across L₁ is 2V.
- Part a: Calculate the voltage across the buzzer.
  - In a series circuit, the sum of the voltages across components equals the total voltage.
  - $V<em>{total} = V</em>{lamp} + V_{buzzer}$
  - $9V = 2V + V_{buzzer}$
  - $V_{buzzer} = 9V - 2V = 7V$
- Part b: Zara adds another lamp, L₂, in series.
  - i. Voltage across L₁: It will decrease because the total voltage is now divided among three components (buzzer, L₁, and L₂).
  - ii. Current in the circuit: The current will decrease because adding more resistance (another lamp) reduces the current, according to Ohm's Law.
Question 4: Marcus's Lamps
- Given: Ten identical lamps rated at 2.0V each, and a 12V battery.
- Part a: Why can't the lamps be connected individually in parallel across the battery?
  - If connected in parallel, each lamp would receive the full 12V, which exceeds their rated voltage of 2.0V, and they would likely burn out.
- Part b:
  - i. Number of lamps in series:
    - To operate properly, the total voltage drop across the lamps should match the battery voltage.
    - $Number : of : Lamps = \frac{Battery : Voltage}{Lamp : Voltage} = \frac{12V}{2V} = 6$
    - Six lamps can be connected in series.
  - ii. Circuit Diagram: A series circuit with the battery and six lamps in a single loop.
  - iii. Voltmeter: Add a voltmeter in parallel across one of the lamps to measure its voltage.

Questions

Question 1: Current through the cell in a parallel circuit.
- Correct Answer: C - The current through the cell is the total of the current in each branch.
Question 2: Identifying parallel circuits.
- Need to visually identify parallel circuits from given diagrams (A, B, C, D).
Question 3: Circuit design with four identical lamps.
- Part a: Draw circuit diagrams.
  - i. Four lamps in series with a cell: Lamps are connected one after another in a single loop.
  - ii. Four lamps in parallel with a cell: Each lamp has its own branch connected directly to the cell.
- Part b: Advantages of parallel connection.
  - If one lamp fails, the others continue to work (independent branches).
  - Each lamp receives the full voltage of the cell, so they all shine brightly.
- Part c: Current calculation in parallel circuit.
  - Given: Total current through the cell is 2.0 A.
  - Since the lamps are identical and in parallel, the current divides equally.
  - $Current : through : each : lamp = \frac{Total : Current}{Number : of : Lamps}$
  - $Current : through : each : lamp = \frac{2.0A}{4} = 0.5A$
Question 4: Calculate the missing current in each circuit. Apply Kirchhoff's current law (total current entering a junction equals total current leaving the junction).
- Circuit A: X = 1 A
- Circuit B: X = 2 A
- Circuit C: X = 0.5 A
- Circuit D: X = 0.5 A
Question 5: Calculate the missing current in each circuit (A, B, C, D)
- Apply Kirchhoff's Current Law to find the missing values (X).
- Circuit A: Total current entering the junction is 6A, and 3A passes through one branch, so X = 6A - 3A = 3A
- Circuit B: 0.2 A + 0.4 A = .6A, so X = .6A
- Circuit C: Current flowing towards the right = .15A + .25A = .4A (same on the other side), so X = .4A
Question 6: Fill in the blanks using the given words (series, parallel, current, voltage).
- An ammeter is used to measure the current through a component.
- An ammeter should be connected in series with the component.
- A voltmeter is used to measure the voltage across a component.
- A voltmeter should be connected in parallel with the component.
Question 7: Identify the correct way to measure current and voltage.
- Correct Answer: Circuit D - Ammeter in series with the lamp and voltmeter in parallel with the lamp.
Question 8: Classroom lamps in parallel.
- Part a: Voltage across one lamp.
  - In a parallel circuit, the voltage across each branch is the same as the supply voltage.
  - Therefore, the voltage across one lamp is 220V.
- Part b: Add a switch to the diagram.
  - The switch should be placed in the main line before the parallel branches to control all lamps simultaneously.
- Part c: Current calculations.
  - i. Current through one lamp:
    - The total current is 2.4 A, and there are six identical lamps.
    - $Current : through : one : lamp = \frac{Total : Current}{Number : of : Lamps} = \frac{2.4A}{6} = 0.4A$
  - ii. Effect of adding two more lamps:
    - The total current from the mains supply will increase because each additional lamp draws its own current from the supply.
Question 9: Arun's series circuit.
- Given: 6V battery, two identical buzzers, and two identical lamps in series; voltage across one buzzer is 1V.
- Part a: Calculate the voltage across one lamp.
  - The total voltage across the buzzers is $2 \times 1V = 2V$ . The total voltage across the lamps is $6V - 2V = 4V$ . The voltage across one lamp is $4V / 2 = 2V$ .
- Part b: Adding another 6V battery in series.
  - i. Voltage across one buzzer: Doubles to 2V because the total voltage doubles and is distributed proportionally.
  - ii. Effect on current: The current in the circuit increases because the total voltage increases (Ohm's Law).

Ohm's Law

$V = IR$ , where V is voltage, I is current, and R is resistance.
1. $I = \frac{V}{R} = \frac{5V}{100\Omega} = 0.05A$
2. $R = \frac{V}{I} = \frac{12V}{0.5A} = 24 \Omega$
3. $V = IR = 0.01A \times 1000\Omega = 10V$
4. $R = \frac{V}{I} = \frac{9V}{0.2A} = 45\Omega$
5. $V = IR = 0.3A \times 7\Omega = 2.1V$
6. Missing resistance.
7. Missing resistance.