Electric Circuits 5 Advanced Circuit Analysis

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# Electric Circuits 5 Advanced Circuit Analysis - PowerPoint PPT Presentation

Electric Circuits 5 Advanced Circuit Analysis. Advanced Circuit Analysis Question 1. Because 1A = 1000mA 5 mA = 0.005 A 150 mA = 0.15 A 0.3 A = 300 mA 0.042 A = 42 mA. ??. ??. ??. ??. Advanced Circuit Analysis Question 2. Because 1k  = 1000  3.5k  = 3500 

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### Electric Circuits 5Advanced Circuit Analysis

Because 1A = 1000mA

• 5 mA = 0.005 A
• 150 mA = 0.15 A
• 0.3 A = 300 mA
• 0.042 A = 42 mA

??

??

??

??

Because 1k= 1000

• 3.5k = 3500
• 25000 = 25 k
• 0.45k = 450
• 22  = 0.22k

??

??

??

??

V

IR

3. If the current through the circuit opposite is 600mA, what is the EMF of the battery?

V = ? I = 600mA= 0.6A R = 20

V = IR

V = 0.6  20

V = 12V

P = VI = I2R = =

4. What power is being dissipated in the resistor opposite?

P = ? V = 6V R = 10

P =

P =

P = 3.6W

5. What is the resistance of a globe rated at 200mW when it is being driven by a 60V battery?

R = ? P = 200mW = 0.2W V = 6V

P =

R =

R=

R = 180

6. What is the power (in mW) being used by an LED that has 0.70V across it and a current of 5.0mA flowing through it?

P = ? V = 0.70V I = 5.0mA = 0.005A

P = VI

P = 0.7  0.05

P = 0.035W

P = 35mW

7. How much heat is dissipated in the resistor opposite in 3 minutes?

W = ? V = 6V R = 100 t = 3  60

= 180s

E =

E =

E = 64.8J

E  65J

8. What is the total resistance of the parallel resistors opposite?

= +

Rtot= 1.5

9. In the circuit opposite:

(a) what is the current in ammeter 1?

I = ? V = 6V R = 16

I

I

I = 0.375A

I  0.38A

(b) what is the current in ammeter 2?

I = ? V = 6V R = 60

I

I

I = 0.10A

16

= +

Rtot= 15

10. Assuming the battery opposite is ideal?

(a) What is the total resistance of the circuit?

Rtot = 2 + 4

Rtot = 6

4

= +

Rtot= 4

10. Assuming the battery opposite is ideal?

(b) What current will flow?

I = ? V = 12V R = 6

I

I

I = 2.0A

4

10. Assuming the battery opposite is ideal?

(c) What will be the power dissipated in the 20 ohm resistor?

P = ? I = 2AR = 2

P = I2R

P = 22 2

P = 8W

4

From q2b

I = 2A

10. Assuming the battery opposite is ideal?

(d) What will be the power drain on the battery?

P = ? I = 2A V = 12V

P = VI

P = 12  2

P = 24W

4

I = 2A

10. Assuming the battery opposite is ideal?

(e) What will be the potential drop across the parallel components?

V = ? I = 2A R = 4V

V = IR

V = 2 4

V = 8V

Or use one of the voltage

divider techniques.

4

I = 2A

From q8f

8V

10. Assuming the battery opposite is ideal?

(f) What current will flow through the 60 ohm resistor?

Method 1 using Vparallel

I = ? V = 8V R = 6

I

I

I = 1.333A

Method 2 –I divider formula

2

4

I = 2A

Method 3 – Ratios

I6: I12

: since I 

12:6

2:1

So I6will be 2 part out of 1 of Itot

2  3  1 = 1.333 A

11. What is Vout?

Method 1 – Every reliable formula

Vout= ?Vin= 16V RD= 6000 Ω Rout= 2000Ω

4.0V

Rout is of the total resistance so

4.0V

Method 3 – Ratios

Rout : RD

2:2

1:3

So Voutwill be 1 part out of 4 of Vin

16  4  1 = 4.0V

2V

12. What is R?

5

5

Method 1 – Every reliable formula

Rout = ? Vin= 12V Vout= 10V RD = 100Ω

500

Method 2 – Ratios on the diagram

Using the ratios on the diagram

R = 100  2 = 500

25V

13. What is R?

5

5

Method 1 – Every reliable formula

RD = ? Vin= 30V Vout= 5.0V Rout = 2000Ω

10000

10k

Method 2 – Ratios on the diagram

Using the ratios on the diagram

R = 2k  5 = 10k

Classical but Long Method

Find R//

= +

Rtot= 1.5

Find V

V = ? I = 12A R = 1.5

V = IR

V = 12  1.5

V = 18V

Find I

I = ? V = 18V R = 6

I = 3.0A

14. What is I6.0?

Method 1 –I divider formula

2

12A

6.0

2.0

### Circuit Design

100V

Circuit DesignQuestion 1

The circuit below has a string of 5V 1W globes.

(a) How many globes are there in the string if the globes are operating to their specifications?

number of globes =

= 20 globes

100V

Circuit DesignQuestion 1

The circuit below has a string of 5V 1W globes.

(b) What is the current flowing in the circuit?

circuit current = current on 1 globe

P = VI

I

I

I = 0.20A

I = ?

V = 5V

P = 1W

Circuit DesignQuestion 2

The circuit for a small set of Christmas tree lights is shown below.

The voltage supply (shown as a DC battery) is 240V. The complete circuit is designed to consume a total of 160W. Which of the following best describes the correct labelling for the lights in the circuit?

• 20W; 30V
• 20W; 240V
• 160W; 30V
• 30W; 20V
Circuit DesignQuestion 3

Arrange the following items into a circuit designed to give a total light intensity of 200W. All items must be connected into the final circuit.

100V

100V

100V, 150W

50V 25W

50V 25W

Circuit DesignQuestion 4

(b) How much current will be flowing through each battery?

Find I in 150W Globe

I = ?

V = 100V

P = 150W

Find I in 25W Globe

I = ?

V = 50V

P = 25W

P = VI

I

I

I = .5A

P = VI

I

I

I = 1.5A

100V

100V, 150W

100V

50V 25W

Total Current = 1.5 + 0.5 = 2.0A

So current through each battery will be 1.0A

0.5A

1.0A

2.0A

1.0A

1.5A

2.0A

50V 25W

Examiners Comment

Series and parallel circuits continue to cause difficulties for some students. Many were unable to work out how much current was flowing in each circuit. It was common for circuit A to have 4 A and circuit B to have 1 A. Other students tried unsuccessfully to obtain the total effective resistance of the circuits

Circuit DesignQuestion 5 – Q18 2011

All the lights in the circuits below are operating normally.Identify which of the circuits, A or B, uses more power and explain why this is the case, using numerical calculationsin your answer?

In Circuit A

Circuit B uses more power

In Circuit B

P = VI

P = 12 × 3

P= 36W

P = VI

P = 12 × 2

P= 24W

P = ?

V = 12V

I = 2.0A

P = ?

V = 12V

I = 3.0A

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Circuit DesignQuestion 5 – Q18 2011

All the lights in the circuits below are operating normally.Identify which of the circuits, A or B, uses more power and explain why this is the case, using numerical calculationsin your answer?

OR

Circuit DesignQuestion 5 – Q18 2011

Examiners Comment

Series and parallel circuits continue to cause difficulties for some students. Many were unable to work out how much current was flowing in each circuit. It was common for circuit A to have 4 A and circuit B to have 1 A. Other students tried unsuccessfully to obtain the total effective resistance of the circuits

All the lights in the circuits below are operating normally.Identify which of the circuits, A or B, uses more power and explain why this is the case, using numerical calculationsin your answer?

55%

Pglobe = VI

P = 12 × 1

P= 12W

Pglobe = VI

P = 6 × 2

P= 12W

Pglobe = ?

V = 12V

I = 1.0A

Pglobe = ?

V = 6V

I = 2.0A

Ptotal = 3 × 12 = 36W

Ptotal = 2 × 12 = 24W

Circuit DesignQuestion 6 – Q 13 2010

Complete the circuit opposite to show the connections when the heater is set to provide a heating power of 600 W.

Note: draw your line in black pen so that it photocopies clearly. Do not use pencil or red pen.

P =

R =

R =

R= 96Ω

Find R required for 600W

R = ?

P = 600

V = 240VRMS

You could have also worked out the power for series and parallel connections to see which gives 600W

For 96Ω the resistors need to be connected in series.

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