section 15.4 series and parallel circuits

8/14/2019 Section 15.4 Series and Parallel Circuits

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Manhattan Press (H.K.) Ltd. 1

Section 15.4Section 15.4

Series and parallel circuitsSeries and parallel circuits

Section 15.4Section 15.4

Series and parallel circuitsSeries and parallel circuits

Parallel circuitParallel circuit

Series circuitSeries circuit

Effects of resistance ofEffects of resistance of

ammeter, voltmeter and cellammeter, voltmeter and cell


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Series and parallel circuits

Bulbs are connected in two ways:

15.4 Series and parallel circuits (SB p. 55)

Go to

Discussion 2Discussion 2

Go to


in series

in parallel


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Series circuit

Series circuit connects electricalcomponents one by one, forming single loop



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total electrical energysupplied by the cell (E)

Series circuit


electrical energy

dissipated inX(E1)

electrical energy

dissipated in Y(E2)

By ,

V= V1 + V2

Q

EV =


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Resistance in series circuit


VVV

III

=+

==

21

21

( )IIIRRR

RIRIIR

VVV

==+=

+=

+=

2121

2211

21

As


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equivalent resistance

R= R1 + R2 + R3 + R4 +

CALWorkshop 5

Simple circuits

(A) Series

circuit
http://cal/E-SimPhy_405.exe


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Drawback of

series circuit

one bulb burns out

whole circuit breakdown


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Drawback of

series circuit

undesirable for domestic wiring


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Parallel circuit

Parallel circuit splits into branches withconnected electrical components



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Resistance in parallel circuit


21 VVV ==

21

21

21

21

21

or

111

As

RR

RRR

RRR

R

V

R

V

R

V

III

+

=

+=

+=

+=


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equivalent resistance

...++++=

4321

11111

RRRRR

CALWorkshop 6

Simple circuits

(B) Parallel

circuit
http://cal/E-SimPhy_405.exe


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Use for domestic wiring

Thinking 2Thinking 2

failure of one bulb

not affect others


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Example 2:Example 2:In the circuit, a battery of voltage of 9 V is connected to a

rheostat. The resistance ofACis 10 and that ofAB is4 .



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

ionLet the current in the circuit beI. The voltage acrossAC

is 9 V. The resistance of the circuit is 10 .

By

I= 0.9 A

I

I

VR

910 =

=



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Example 2: (Cont)Example 2: (Cont)(b) What is the voltage acrossAB?

(c) What is the voltage across BC?

Solut

ionThe current flowing throughAB is 0.9 A. The resistance

is 4 .

By V =IR

V = 0.9 4 = 3.6 V


Solut

ionVoltage acrossBC= 9 3.6 = 5.4 V


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Example 3:Example 3:In the circuit as shown, find

(a) the total resistance of the resistors

Solution



Resistance of the 3- and 6- resistors = = 2

Total resistance = 2 + 4 = 6 63

63

+


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Example 3: (Cont)Example 3: (Cont)(b) find the current flowing through the 4- resistor,

(c) find the currents flowing through the 3- and 6- resistors,

Solut

ionCurrent through the 4- resistor = = 0.5 A6

3=

R

V


Solut

ion

Voltage across the 3- and 6- resistors =IR = 0.5 2 = 1

V

Current flowing through the 3- resistor

Current flowing through the 6- resistor

A3

1==

R

V

A6

1==

R

V


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

SolutionVoltage across the 4- resistor = IR = 0.5 4 = 2 V



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Class Practice 3:Class Practice 3:In the circuit, a 100- and a 300- resistors are

connected in parallel with a cell of 20 V.

(a) Find the total resistance of the resistors.

________________

Total resistance = _________________

Ans

wer

75


=+=

21

111

RRR 300

1

100

1+


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

Current drawn from the cell (I) =

______________

(c) Find the current passing through each resistor.

Current passing through 100- resistor

_________________

Current passing through 300- resistor

_________________

Ans

wer


=

R

VA27.0

75

20=

A2.0100

20==

R

V

A07.0300

20==

R

VAns

wer

Ans

wer


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

Ans

wer



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Class Practice 4: (Cont)Class Practice 4: (Cont)


Since the 100- and 200- resistors

are connected in series.

Total resistance of the 100- and 200- resistors

= 100 + 200 = 300 The 300- resistor connected with them in parallel, by

Total resistance of the 100- , 200- and 300- resistors = 150

300

1

300

11+=

R


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

The current drawn from the cell

________________

(c) Find the currents flowing through each of the resistors.

Ans

wer


A2.0150

30==

R

V

Ans

werThe current flowing through the 100- and 200- resistors

The current flowing through the 300- resistor

A1.0

300

30===

R

V

A1.0300

30===

R

V


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

Ans

wer

Voltage across the 100- resistor

=IR = 0.1 100 = 10 V

Voltage across the 200- resistor

=IR = 0.1 200 = 20 V

Voltage across the 300- resistor= 30 V



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Effects of resistance of ammeter,

voltmeter and cell


They have resistance too

effect on current / voltage

Go to


battery

ammeter

voltmeter


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Effects of resistance of ammeter


( few ohms )

Small resistance circuit

Total resistance= Resistance ofR1+ Resistance of ammeter (RA)

comparable total resistance


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resistance


I & V of R1


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Total resistance= Resistance ofR2+ Resistance of ammeter (RA)

Large resistance circuit

R2 >>RAtotal resistanceremain unchanged

I & V of R2

remain

unchanged

( few ohms )


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Effects of resistance of voltmeter


Total current= Current through R1+ Current through voltmeter


RV >>R1total currentremain unchanged

( few hundred ohms )


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Large resistance circuit

Total current= Current through R1+ Current through voltmeter

RV &R1 comparable total current

Effects of resistance of voltmeter

( few hundred ohms )

i d ll l i i ( )


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internal resistance (r) (few ohms)

Voltage of the cell =I

(R

+r

)

affects V&Iofsmall resistance circuit only

Effect of resistance of cell



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To section 15.5

15 4 S i d ll l i it (SB 55)
http://e-ch15_05.ppt/http://e-ch15_05.ppt/


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Discussion 2:Discussion 2:

When one of the light bulbs on a Christmas tree isremoved, what happen to the other light bulbs? Why?

The light bulbs in 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, theother light bulbs in different strings

are connected in parallel. Therefore,

they still light up.

Ans

wer


Return to

TextText

15 4 S i d ll l i it (SB 56)


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You are given two resistors, a dry cell and someconnecting 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.


15 4 S i d ll l i it (SB 56)


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Discussion 3:Discussion 3:(Cont)(Cont)

(Hint: Imagine the electric circuit as a water pipesystem 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.)

(a) In series

Ans

wer


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

(a)(i)

15 4 Series and parallel circuits (SB p 56)


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Discussion 3:Discussion 3:(Cont)(Cont)

(b) In parallel

Ans

wer


Return to

TextText

(ii)I1 =I2 +I3

(iii) V1 = V2 = V3

(b)(i)



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Injury caused by an electric shock depends onthe amount of the current that flows in the

body. Why do we see signs that read

"Danger - High Voltage" rather than

"Danger - High Current"? Answer

This is because the current depends on

the resistance of the body. The voltage

is fixed by the power source. Therefore awarning of high voltage is more relevant

than that of high current.Return to

TextText




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In the circuit, when the switch is closed, whathappen to the readings of ammetersA1,A2

and voltmeterV1?


Ans

wer



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Thinking 3 (Cont)Thinking 3 (Cont)

The reading ofV1 remains unchanged because it is

equal to the voltage of the cell. The voltage acrossR1 is

equal to the voltage of the cell. ByI

= , the reading ofA1 remains unchanged.

The reading ofA2 is equal to the sum of the current

passing throughR1 andR2. If the switch is closed, there

is a current passing throughR2. Therefore, the reading

ofA2 increases.

Return to

TextText


R

V



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1. In CircuitA, find the voltages across theresistor (R) and currents passing through it if the

resistance ofRis

(a) 0.1 , and

(b) 1 k .


Ans

wer

(a) VA1 = 3 V

IA1 = = 30 A1.0

3

(b) VA2 = 3 V

IA2 = = 3 103 A

k1

3

CircuitA



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Discussion 4: (Cont)Discussion 4: (Cont)2. In Circuit B, find the voltages across the resistor

(R) if the resistance ofRis

(a) 0.1 , and

(b) 1 k .


Ans

wer

IB1 = = 0.588 A

VB1

= 0.588 0.1

= 0.0588 V

51.0

3

+

(a)IB2 = = 2.985 10

3A

VB2

= 2.985 103 1 k

= 2.985 V

5k1

3

+

(b)

CircuitB



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Discussion 4: (Cont)Discussion 4: (Cont)3. In Circuit B, if we take the 5- resistor away,

what happen to the voltage and the current in resistor

R?


Ans

wer

CircuitB becomesCircuitA.

(i) If the resistor ofR is

small (e.g., 0.1 ),

IA1 >>IB1VA1 >>VB1

(ii) If the resistance ofR is

large (e.g., 1 k ),

IA2 IB2VA2 VB2

CircuitB



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Discussion 4: (Cont)Discussion 4: (Cont)

44. In Circuit C, find the total currents flowing out ofthe battery if the resistance ofRis(a) 0.1 , and

(b) 1 k .


Ans

wer

(a) IC1 =

= 30.000 A

k103

1.03 + (b) IC2 =

= 3.3 103 A

k103

k13 +

Circuit C



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Discussion 4: (Cont)Discussion 4: (Cont)5. In Circuit C, if we take the 10- k resistor away,

what happens to the current flowing through resistor

R?

p ( p )

Circuit Calso becomes

CircuitA.

(i) If the resistor ofR is

small (e.g., 0.1

),IA1 IC1

(ii) If the resistance ofR is

large (e.g., 1 k

),IA2


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The resistance of a resistor can be found byusing either circuitA or circuit B shown below.

Explain which circuit is suitable for finding

1. high resistance and

2. low resistance.Ans

wer

p ( p )

R R

Circuit BCircuitA



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1. CircuitAVoltmeter reading = VacrossR + Vacross ammeter

SinceRresistor >>Rammeter and VR,

VacrossR >> Vacross ammeter.So VacrossR ~ voltmeter reading.

Ammeter reading =IthroughR

Therefore, the resistance calculatedfromR = V/Iis close to the actual value.

p ( p )

R

R15.4 Series and parallel circuits (SB p. 66)


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2. CircuitB

Ammeter reading =IthroughR +Ithrough voltmeter

SinceRresistor Ithrough voltmeter.

ThereforeIthroughR ~ ammeter reading.

Voltmeter reading = VacrossR

Therefore, the resistance calculated

fromR = V/Iis close to the actual value.

In an ideal case, we should always use an ammeter with a very low

resistance and a voltmeter with an extremely high resistance. Then,

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TextText

R

section 15.4 series and parallel circuits

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