Section 15.4 Series And Parallel Circuits

Section 15.4 Series and parallel circuits • Series circuit • Parallel circuit • Effects of resistance of ammeter, voltmeter and cell © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 55)

Series and parallel circuits Bulbs are connected in two ways: in series in parallel

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Discussion 2

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Discussion 3

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15.4 Series and parallel circuits (SB p. 56)

Series circuit Series circuit — connects electrical components one by one, forming single loop

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15.4 Series and parallel circuits (SB p. 57)

Series circuit total electrical energy supplied by the cell (E) electrical energy dissipated in X (E1)

electrical energy dissipated in Y (E2) By V = E , Q V = V1 + V 2

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15.4 Series and parallel circuits (SB p. 57)

Resistance in series circuit

I1 = I2 = I V1 + V2 = V As V = V1 + V2 IR = I1R1 + I2R2

R = R1 + R2 ( I 1= I2 = I )

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15.4 Series and parallel circuits (SB p. 57)

Resistance in series circuit

CAL Workshop 5

equivalent resistance

Simple circuits

R = R 1 + R 2 + R3 + R4 + …

(A) Series circuit © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 58)

Resistance in series circuit Drawback of series circuit

one bulb burns out → whole circuit breakdown © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 58)

Resistance in series circuit Drawback of series circuit undesirable for domestic wiring

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15.4 Series and parallel circuits (SB p. 58)

Parallel circuit Parallel circuit — splits into branches with connected electrical components

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15.4 Series and parallel circuits (SB p. 59)

Resistance in parallel circuit

V = V1 = V2 As

I = I1 + I2 V V V = + R R1 R2 1 1 1 = + R R1 R2

or © Manhattan Press (H.K.) Ltd.

R1R2 R= R1 + R2 10

15.4 Series and parallel circuits (SB p. 59)

Resistance in parallel circuit

CAL Workshop 6 Simple circuits (B) Parallel circuit © Manhattan Press (H.K.) Ltd.

equivalent resistance

1 1 1 1 1 = + + + + ... R R1 R2 R3 R4 11

15.4 Series and parallel circuits (SB p. 60)

Resistance in parallel circuit Use for domestic wiring

Thinking 2 © Manhattan Press (H.K.) Ltd.

failure of one bulb → not affect others 12

15.4 Series and parallel circuits (SB p. 60)

Example 2: In the circuit, a battery of voltage of 9 V is connected to a rheostat. The resistance of AC is 10 Ω and that of AB is 4Ω.

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15.4 Series and parallel circuits (SB p. 60)

Example 2: (Cont) (a) What is the current in the circuit?

Solut ion Let the current in the circuit be I. The voltage across AC is 9 V. The resistance of the circuit is 10 Ω . By R = V I 9 10 = I I = 0.9 A

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15.4 Series and parallel circuits (SB p. 60)

Example 2: (Cont)

Solut (b) What is the voltage across AB? ion The current flowing through AB is 0.9 A. The resistance is 4 Ω . By V = IR V = 0.9 × 4 = 3.6 V (c) What is the voltage across BC? Solut Voltage across BC = 9 − 3.6 = 5.4ion V

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15.4 Series and parallel circuits (SB p. 61)

Example 3: In the circuit as shown, find

Solut ion

(a) the total resistance of the resistors

3× 6 Resistance of the 3-Ω and 6-Ω resistors = 3+6 Total resistance = 2 + 4 = 6 Ω

=2Ω Thinking 3

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15.4 Series and parallel circuits (SB p. 61)

Example 3: (Cont) (b) find the current flowing through the 4-Ω resistor, Solut V 3 Current through the 4-Ω resistor = = = 0.5 A ion R 6 (c) find the currents flowing through the 3-Ω and 6-Ω resistors, Voltage across the 3-Ω and 6-Ω resistors = IR = 0.5 × 2 = 1 V Solut V 1 flowing through the 3-Ω resistor Current = = A R 3

Current V 1 flowing through the 6-Ω resistor = = A R 6

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ion

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15.4 Series and parallel circuits (SB p. 61)

Example 3: (Cont) (d) find the voltage across the 4-Ω resistor.

Solut Voltage across the 4-Ω resistor = IR = 0.5 × 4 = ion 2V

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15.4 Series and parallel circuits (SB p. 61)

Class Practice 3: In the circuit, a 100-Ω and a 300-Ω resistors are connected in parallel with a cell of 20 V.

(a) Find the total resistance of the resistors. 1 1 1 1 1 + = + = ________________ 100 300 R R1 R2

Ans wer

75 Ω Total resistance = _________________ © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 62)

Class Practice 3: (Cont) (b) Find the current drawn from the cell.

V Current drawn from the cell (I) = = R

20 Ans = 0.27 A 75 wer

______________ (c) Find the current passing through each resistor. Ans Current passing through 100-Ω resistor V 20 wer = = 0.2 A R 100 _________________ Current passing through 300-Ω resistor V 20 = = 0.07 A R 300 _________________

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Ans wer 20

15.4 Series and parallel circuits (SB p. 62)

Class Practice 4: (a) In the circuit, find the total resistance.

Ans wer

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15.4 Series and parallel circuits (SB p. 62)

Class Practice 4: (Cont) Since the 100-Ω and 200-Ω resistors are connected in series. Total resistance of the 100-Ω and 200-Ω resistors = 100 Ω + 200 Ω = 300 Ω The 300-Ω resistor connected with them in parallel, by 1 1 1 = + R 300 300 Total resistance of the 100-Ω , 200-Ω and 300-Ω resistors = 150 Ω

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15.4 Series and parallel circuits (SB p. 62)

Class Practice 4: (Cont) (b) Find the total current drawn from the cell. The current drawn from the cell Ans

V 30 = = 0.2 A ________________ R 150

wer

(c) Find the currents flowing through each of the resistors. Ans wer The current flowing through the 100-Ω and 200-Ω resistors =

V 30 = = 0.1 A R 300

The current flowing through the 300-Ω resistor = © Manhattan Press (H.K.) Ltd.

V 30 = = 0.1 A R 300 23

15.4 Series and parallel circuits (SB p. 63)

Class Practice 4: (Cont) (d) Find the voltage across each of the resistors.

Ans wer

Voltage across the 100-Ω resistor = IR = 0.1 × 100 = 10 V Voltage across the 200-Ω resistor = IR = 0.1 × 200 = 20 V Voltage across the 300-Ω resistor = 30 V

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15.4 Series and parallel circuits (SB p. 63)

Effects of resistance of ammeter, voltmeter and cell

battery

voltmeter

ammeter Go to

Discussion 4 © Manhattan Press (H.K.) Ltd.

They have resistance too → effect on current / voltage 25

15.4 Series and parallel circuits (SB p. 64)

Effects of resistance of ammeter Small resistance circuit

Total resistance ( few ohms ) = Resistance of R1 + Resistance of ammeter (RA) comparable → total resistance ↑ © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 64)

Effects of resistance of ammeter Small resistance circuit resistance ↑ → I & V of R1 ↓

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15.4 Series and parallel circuits (SB p. 64)

Effects of resistance of ammeter Large resistance circuit → I & V of R2 remain unchanged Total resistance ( few ohms ) = Resistance of R2 + Resistance of ammeter (RA) R2 >> RA → total resistance remain unchanged © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 65)

Effects of resistance of voltmeter Small resistance circuit

Total current ( few hundred ohms ) = Current through R1 + Current through voltmeter RV >> R1 → total current remain unchanged © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 65)

Effects of resistance of voltmeter Large resistance circuit

Total current ( few hundred ohms ) = Current through R1 + Current through voltmeter RV & R1 comparable → total current ↑ © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 66)

Effect of resistance of cell

internal resistance (r) (few ohms) Voltage of the cell = I ( R + r ) ∴ affects V & I of small resistance circuit only Thinking 4 © Manhattan Press (H.K.) Ltd.

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To section 15.5

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15.4 Series and parallel circuits (SB p. 55)

Discussion 2: When one of the light bulbs on a Christmas tree is removed, what happen to the other light bulbs? Why? Ans bulbs in wer

The light the affected string do not light up but the light bulbs in other strings do. This is because the unlit bulbs are connected in series. If one light bulb is failed, the other bulbs do not light up. However, the other light bulbs in different strings are connected in parallel. Therefore, they still light up. © Manhattan Press (H.K.) Ltd.

Return to

Text

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15.4 Series and parallel circuits (SB p. 56)

Discussion 3: You are given two resistors, a dry cell and some connecting wires. Connect the resistors (a) in series and (b) in parallel with the dry cell to form a closed circuit. In each case, (i) draw the circuit diagram, (ii) find the relation between the currents passing through the dry cell and the two resistors, and (iii) find the relation between the voltages across the dry cell and the two resistors. © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 56)

Discussion 3: (Cont) (Hint: Imagine the electric circuit as a water pipe system with the dry cell acting as a pump, the connecting wires as pipes, the current as a flow of water and resistors as water turbines.) Ans wer (a) In series (a)(i)

(ii) I1 = I2 = I3 (iii) V1 = V2 + V3

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15.4 Series and parallel circuits (SB p. 56)

Discussion 3: (Cont)

(b) In parallel

Ans wer

(b)(i)

(ii) I1 = I2 + I3 (iii) V1 = V2 = V3 © Manhattan Press (H.K.) Ltd.

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Text 36

15.4 Series and parallel circuits (SB p. 60)

Thinking 2

Injury caused by an electric shock depends on the amount of the current that flows in the body. Why do we see signs that read "Danger - High Voltage" rather than Ans "Danger - High Current"? wer This is because the current depends on the resistance of the body. The voltage is fixed by the power source. Therefore a warning of high voltage is more relevant Return to than that of high current. Text © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 61)

Thinking 3

In the circuit, when the switch is closed, what happen to the readings of ammeters A1, A2 Ans and voltmeter V1? wer

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15.4 Series and parallel circuits (SB p. 61)

Thinking 3 (Cont) The reading of V1 remains unchanged because it is equal to the voltage of the cell. The voltage across R1 is V equal to the voltage of the cell. By I = R , the reading of A1 remains unchanged. The reading of A2 is equal to the sum of the current passing through R1 and R2. If the switch is closed, there is a current passing through R2. Therefore, the reading Return to

of A increases.

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Text

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15.4 Series and parallel circuits (SB p. 63)

Discussion 4:

1. In Circuit A, find the voltages across the resistor (R) and currents passing through it if the resistance of R is Circuit A (a) 0.1 Ω , and (b) 1 kΩ . Ans wer

(a) VA1 = 3 V 3 IA1 = 0.1 = 30 A © Manhattan Press (H.K.) Ltd.

(b) VA2 = 3 V 3 IA2 = 1 k = 3 × 10−3 A 40

15.4 Series and parallel circuits (SB p. 63)

Discussion 4: (Cont) 2. In Circuit B, find the voltages across the resistor (R) if the resistance of R is Circuit B (a) 0.1 Ω , and Ans (b) 1 kΩ . wer (a)

3 IB1 = = 0.588 A 0.1 + 5 VB1 = 0.588 × 0.1 = 0.0588 V

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(b) IB2 =

3 = 2.985 × 10−3A 1k +5

VB2 = 2.985 × 10−3 × 1 k = 2.985 V

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15.4 Series and parallel circuits (SB p. 63)

Discussion 4: (Cont) 3. In Circuit B, if we take the 5- Ω resistor away, what happen to the voltage and the current in resistor Circuit B R? Ans wer

Circuit B becomes Circuit A. (i) If the resistor of R is small (e.g., 0.1 Ω ), IA1 >> IB1 VA1 >> VB1 © Manhattan Press (H.K.) Ltd.

(ii) If the resistance of R is large (e.g., 1 k Ω ), IA2 ≅ IB2 VA2 ≅ VB2 42

15.4 Series and parallel circuits (SB p. 64)

Discussion 4: (Cont) 4. In Circuit C, find the total currents flowing out of the battery if the resistance of R is Circuit C (a) 0.1 Ω , and Ans (b) 1 kΩ . wer 3 3 3 3 + + (a) IC1 = (b) IC2 = 0.1 10 k 1 k 10 k = 30.000 A

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= 3.3 × 10−3 A

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15.4 Series and parallel circuits (SB p. 64)

Discussion 4: (Cont) 5. In Circuit C, if we take the 10- kΩ resistor away, what happens to the current flowing through resistor R? Circuit C Circuit C also becomes Circuit A. (i) If the resistor of R is small (e.g., 0.1 Ω ), IA1 ≈ IC1

Ans wer

(ii) If the resistance of R is large (e.g., 1 k Ω ), IA2 < IC2 Return to

Text © Manhattan Press (H.K.) Ltd.

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15.4 Series and parallel circuits (SB p. 66)

Thinking 4

The resistance of a resistor can be found by using either circuit A or circuit B shown below. Explain which circuit is suitable for finding 1. high resistance and Ans wer 2. low resistance. Circuit A

Circuit B R

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R

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15.4 Series and parallel circuits (SB p. 66)

Thinking 4 (Cont) 1. Circuit A Voltmeter reading = V across R + V across ammeter Since Rresistor >> Rammeter and V ∝ R, V across R >> V across ammeter. So V across R ~ voltmeter reading. Ammeter reading = I through R Therefore, the resistance calculated from R = V / I is close to the actual value.

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R

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15.4 Series and parallel circuits (SB p. 66)

Return to

Thinking 4 (Cont) Text 2. Circuit B Ammeter reading = I through R + I through voltmeter Since Rresistor << Rvoltmeter and I ∝ 1/R, I through R >> I through voltmeter. Therefore I through R ~ ammeter reading. Voltmeter reading = V across R Therefore, the resistance calculated from R = V / I is close to the actual value.

R

In an ideal case, we should always use an ammeter with a very low resistance and a voltmeter with an extremely high resistance. Then, both circuits make no major difference for measuring the resistance. © Manhattan Press (H.K.) Ltd. 47