Electric Circuits 5 Advanced Circuit Analysis

Electric Circuits 5Advanced Circuit Analysis

Advanced Circuit AnalysisQuestion 1

Because 1A = 1000mA

i. 5 mA = 0.005 Aii. 150 mA = 0.15 Aiii.0.3 A = 300 mAiv. 0.042 A = 42 mA

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??

??

??

Advanced Circuit AnalysisQuestion 2

Because 1k = 1000

i. 3.5k = 3500 ii. 25000 = 25 kiii.0.45k = 450 iv. 22 = 0.22 k

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Advanced Circuit Analysis

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 = IRV = 0.6 20V = 12V

V

I R

P = VI = I2R = =


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

P = ? V = 6V R = 10P = P = P = 3.6W


5. What is the resistance of a globe rated at 200mW when it is being driven by a 60V 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 0.05P = 0.035WP = 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 = 64.8JE 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 = 16I I I = 0.375AI 0.38A

(b) what is the current in ammeter 2?I = ? V = 6V R = 60

I I I = 0.10A

= + Rtot = 15

16


10. Assuming the battery opposite is ideal? (a) What is the total resistance of the circuit?

Rtot = 2 + 4Rtot = 6

= + Rtot = 4

4


10. Assuming the battery opposite is ideal? (b) What current will flow?

I = ? V = 12V R = 6I I I = 2.0A

4


10. Assuming the battery opposite is ideal? (c) What will be the power dissipated in the 20

ohm resistor?P = ? I = 2A R = 2

P = I2RP = 22 2P = 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 = VIP = 12 2P = 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 = IRV = 2 4V = 8V

Or use one of the voltagedivider techniques.

4

I = 2A


10. Assuming the battery opposite is ideal? (f) What current will flow through the 60 ohm resistor? Method 1 using Vparallel

I = ? V = 8V R = 6I I I = 1.333A

Method 2 –I divider formula

2

4

I = 2A From q8f8V

Method 3 – Ratios I6 : I12

: since I 12:6 2:1So I6 will 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

Method 2 – Formula in your headRout is of the total resistance so

4.0V

Method 3 – RatiosRout : RD

2:2 1:3So Vout will be 1 part out of 4 of Vin

16 4 1 = 4.0V


12. What is R?

Method 1 – Every reliable formulaRout = ? Vin

= 12V Vout = 10V RD = 100 Ω

500

Method 2 – Ratios on the diagramUsing the ratios on the diagram R = 100 2 = 500

2V

5

5


13. What is R?

Method 1 – Every reliable formulaRD = ? Vin

= 30V Vout = 5.0V Rout = 2000 Ω

10000 10k

Method 2 – Ratios on the diagramUsing the ratios on the diagram R = 2k 5 = 10k

25V

5

5


14. What is I6.0?

Method 1 –I divider formula 2

2.0

6.012A

Classical but Long MethodFind R//

= + Rtot = 1.5

Find VV = ? I = 12A R = 1.5V = IRV = 12 1.5V = 18V

Find II = ? V = 18V R = 6 I = 3.0A

Circuit Design

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

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

circuit current = current on 1 globe

100V

I = ?V = 5VP = 1W

P = VIIII = 0.20A

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?

A. 20W; 30VB. 20W; 240VC. 160W; 30VD. 30W; 20V

Circuit DesignQuestion 3Arrange 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

50V 25W

100V, 150W

50V 25W

Circuit DesignQuestion 4(b) How much current will be flowing through each battery?

100V

100V

50V 25W

100V, 150W

50V 25W

Find I in 150W GlobeI = ?V = 100VP = 150WP = VIIII = 1.5A

Find I in 25W GlobeI = ?V = 50VP = 25WP = VIIII = .5A

Total Current = 1.5 + 0.5 = 2.0ASo current through each battery will be 1.0A

1.5A

0.5A2.0A

2.0A1.0A

1.0A

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 calculations in your answer?

P = VIP = 12 × 2P= 24W

P = ?V = 12VI = 2.0A

P = VIP = 12 × 3P= 36W

P = ?V = 12VI = 3.0A

Circuit B uses more power

In Circuit A In Circuit B

Examiners CommentSeries 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

55%



OR



Pglobe = VIP = 6 × 2P= 12W

Pglobe = ?V = 6VI = 2.0A

Pglobe = VIP = 12 × 1P= 12W

Pglobe = ?V = 12VI = 1.0A

Ptotal = 2 × 12 = 24W

55%

Ptotal = 3 × 12 = 36W

Examiners CommentSeries 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 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.

Find R required for 600W

R = ?P = 600V = 240VRMS

P = R = R = R = 96Ω

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

Note: draw your line in black pen so that it

photocopies clearly. Do not use pencil or red pen.

58%

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

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Electric Circuits 5 Advanced Circuit Analysis