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W = V 2 t/R R = V 2 t/E 115 W .

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Do Now: A appliance uses 450 kJ of energy over 1 hour when connected with a p.d. of 120-V. What is the resistance of the appliance?. W = V 2 t/R R = V 2 t/E 115 W. Sketching with Circuit Symbols. To represent components on circuit use symbols. See ref tables. Circuit Symbols. - PowerPoint PPT Presentation
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Do Now: An appliance uses 450 kJ of energy over 1 hour when connected with a p.d. of 120-V. What is the resistance of the appliance? W = V 2 t/R R = V 2 t/E R = (120V) 2 (3600 s) / 450x10 3 J 115 .
Transcript
Page 1: W = V 2 t/R  R = V 2 t/E 115  W .

Do Now: An appliance uses 450 kJ of energy over 1 hour when connected with a p.d. of 120-V. What is the resistance of the appliance?

• W = V2t/R

• R = V2t/E

• R = (120V)2(3600 s) / 450x103 J

• 115 .

Page 2: W = V 2 t/R  R = V 2 t/E 115  W .

To represent components on circuit use symbols.

See ref tables

Sketching with Circuit Symbols

Page 3: W = V 2 t/R  R = V 2 t/E 115  W .

CircuitSymbols

Page 4: W = V 2 t/R  R = V 2 t/E 115  W .

1. Use your tables to sketch the circuit below with the appropriate

symbols

Page 5: W = V 2 t/R  R = V 2 t/E 115  W .

2 Types of Circuits

• Series – single pathway for current flow. Components connected in succession like a chain.

• Parallel – pathway with more than 1 branch – wire splits.

Page 6: W = V 2 t/R  R = V 2 t/E 115  W .

Analyzing Components onElectric Circuits

Each component on a circuit has its own current, voltage, resistance.

How can those variables be determined for individual resistors in circuits?

Page 7: W = V 2 t/R  R = V 2 t/E 115  W .

Use Ohm’s Law for individual components:

I1 = V1/R1 V1 = I1R1 or R1 = V1/I1

Or to analyze total current, voltage resistance for the entire circuit.Itot = Vtot/Req

Vtot = ItotReq Req = Vtot/Itot

Page 8: W = V 2 t/R  R = V 2 t/E 115  W .

The power supply (battery) provides the total E, the components may share (divide) the total E (p.d.), current, or both.

Page 9: W = V 2 t/R  R = V 2 t/E 115  W .

Analyzing Series CircuitsSeries circuits are a chain of components connected in a circle. Charges have 1 path.

Page 10: W = V 2 t/R  R = V 2 t/E 115  W .

Because of the conservation of charge, & because there is only one route, the current (I) is the same through every component.

I1 = I2 = I3 = I4 …

Page 11: W = V 2 t/R  R = V 2 t/E 115  W .

2. the reading on ammeter 1 is 3-A, what is the reading on ammeter 4?

1. 4.

2. 3.

Page 12: W = V 2 t/R  R = V 2 t/E 115  W .

Resistance

As more resistors are added in a series connection, the total equivalent resistance of

the circuit increases.

equivalent resistance is the simple addition of each resistor on the circuit.

R1 + R2+ R3 = Req.

Page 13: W = V 2 t/R  R = V 2 t/E 115  W .

3. Given the circuit:

2-. 4-. 6-.

What is the equivalent resistance?

Page 14: W = V 2 t/R  R = V 2 t/E 115  W .

Voltage on a series circuit.

The battery, or generator or source provides the Vtot for the circuit.

The Vtot is the addition of each p.d. across each resistor on the circuit.

V1 + V2+ V3 ~ Vtot.= the battery voltage

Page 15: W = V 2 t/R  R = V 2 t/E 115  W .

4. Given the circuit:

1-V. 2-V. 3-V.

What is the total circuit voltage?What is the battery voltage?

6V

Page 16: W = V 2 t/R  R = V 2 t/E 115  W .

5. The p.d. or V across each resistor can be found:

V1 = IR1. V2 = IR2. V3 = IR3.

• If resistance on V3 is 3 and the current is 1 A, what is the voltage V3?

• What is the current through bulb 1?

 • What will happen if one of the bulbs burns out?

V = IR = (3)(1A) = 3V

1A

Page 17: W = V 2 t/R  R = V 2 t/E 115  W .

Since there is only one path for the charges to follow, if one conductor (resistor) is disconnected, the circuit is broken. The current flow stops. The bulbs go out.

Page 18: W = V 2 t/R  R = V 2 t/E 115  W .

Since the devices must share the voltage, as more are added the energy of the charges decreases. The bulbs become dimmer.

Page 19: W = V 2 t/R  R = V 2 t/E 115  W .

• Read txt 730 – 739 do 739 #1, 2, 4, 5

• For problem 1,2 above sketch the circuit along side of the calculations w proper symbols for problem solving.

In Class tx pg 745 #4a, 4c.

Page 20: W = V 2 t/R  R = V 2 t/E 115  W .

Parallel Circuits

Page 21: W = V 2 t/R  R = V 2 t/E 115  W .

The current reaches a fork, can choose path.

What’s Different?

Page 22: W = V 2 t/R  R = V 2 t/E 115  W .

Through which resistor does more charge flow? Why?

Page 23: W = V 2 t/R  R = V 2 t/E 115  W .

Sketching: Make rungs or branches.

Page 24: W = V 2 t/R  R = V 2 t/E 115  W .

• Resistors all connected across the V , all V equal to the battery voltage.

• V1 = V2 =V3 = Vtot

• Current (I), charges reach a junction, they divide.

• Itot in the circuit = currents in each branch.

• Itot = I1 + I2 + I3 …

• Individual currents

• I1 = VR1

Page 25: W = V 2 t/R  R = V 2 t/E 115  W .

Resistance goes down as more branches are added:

1/Req = 1/R1 + 1/R2 + 1/R3 …

Where Req is the equivalent or total resistance.

Page 26: W = V 2 t/R  R = V 2 t/E 115  W .

*Because resistance is a reciprocal relationship, the Req must be less than the smallest resistance on the circuit. As you add resistors, the resistance goes down!!

Page 27: W = V 2 t/R  R = V 2 t/E 115  W .

Since parallel circuits offer more than one path for the charge to flow, individual parts can be disconnected and charge will still flow through other branches.

Page 28: W = V 2 t/R  R = V 2 t/E 115  W .

Since the voltage is equal on each branch, adding more branches does not reduce the energy each branch receives. Add more bulbs, the others stay bright!

Page 29: W = V 2 t/R  R = V 2 t/E 115  W .

Ex 1: A 9V battery is connected in series to 2 bulbs: 4, & 2.

A) Sketch the diagram with the proper symbols. Show real and conventional current flow direction.

B) Find the equivalent or total resistance on the circuit.

C) Find the total current in the circuit. D) What is the voltage in each branch?E) Find the current in each branch.

Page 30: W = V 2 t/R  R = V 2 t/E 115  W .

Ex 2: A 9V battery is connected in parallel to 2 bulbs: 4, & 2.

A) Sketch the diagram with the proper symbols. Show real and conventional current flow direction.

B) Find the equivalent or total resistance on the circuit.

C) Find the total current in the circuit. D) What is the voltage in each branch?E) Find the current in each branch.

Page 31: W = V 2 t/R  R = V 2 t/E 115  W .

F) Add the currents from each branch together. How do they relate to the total current?

G) Now add a 3 bulb to the circuit. Recalculate the equivalent resistance.

H) How does the new resistance on the circuit compare to the original?

I) Recalculate the power of each bulb.

J) What happened to the bulbs’ brightness when the 3 bulb was added?

Page 32: W = V 2 t/R  R = V 2 t/E 115  W .

Summery

Parallel. Add resistance, total

• Itot goes up.

• Req goes down.

• Indiv. P/ Brightness equal.

• Lowest R = most P brightest.

• Remove 1, others stay on same power/brightness

Series. Add resistance, total

• Itot goes down

• Req goes up.

• Indiv P/brightness decrease.

• Highest R = bright/most P

• Remove 1, others go out.

• Less bulbs each is brighter.

Page 33: W = V 2 t/R  R = V 2 t/E 115  W .

Hwk:Text pg 745 #2 - 5.

Sketch the circuits in #4, 5.

Page 34: W = V 2 t/R  R = V 2 t/E 115  W .

Power = Brightness

The bulb brightness in each resistor is dependent on the power in each resistor.

P = VI P = I2R P = V2/R,

can calculate the power and deduce the brightness.

Page 35: W = V 2 t/R  R = V 2 t/E 115  W .

Which equation is more useful for:

• Series?

• Parallel?

P = I2R

P = V2

R

Page 36: W = V 2 t/R  R = V 2 t/E 115  W .

Ex 2: 2 bulbs in series connected to 6V cell. R1 = 2, R2 = 4 .

Find the power in each.

What is relative brightness?

What if the 2 is removed? What happens to the brightness of the remaining bulbs?

Page 37: W = V 2 t/R  R = V 2 t/E 115  W .

Ex 3: 3 bulbs in parallel connected to 6V cell. R1 = 2, R2 = 3 , R3 = 4.

There are 2 voltmeters: 1 reads the 4 & 1 reads the 2 bulb.

2 ammeters - 1 reads the total current 1 reads the current in the 3 branch.

A. Sketch the circuit. Find the power in each.

B. What is relative brightness?

Page 38: W = V 2 t/R  R = V 2 t/E 115  W .

C. Add a 100 bulb to the circuit & calculate the power in each. What happens to the brightness? What happens to the total current?

D. What if the 2 is removed? What happens to the brightness of the remaining bulbs?

Page 39: W = V 2 t/R  R = V 2 t/E 115  W .

Use of Meters

Page 40: W = V 2 t/R  R = V 2 t/E 115  W .

Ammeters measure current so circuit current must flow through meter.

Connect meter in series to measure current.

Page 41: W = V 2 t/R  R = V 2 t/E 115  W .

Ideal ammeter has zero resistance.

• What do you think will happen if an ammeter is connected in parallel by accident?

Page 42: W = V 2 t/R  R = V 2 t/E 115  W .

Voltmeters measure p.d. across resistors so must be connected in parallel.

Page 43: W = V 2 t/R  R = V 2 t/E 115  W .

Ideal voltmeter has infinite resistance.

• What do you think will happen if a voltmeter is connected in series by accident?

Page 44: W = V 2 t/R  R = V 2 t/E 115  W .

Kirchoff’s Laws The current entering junction = current exiting. Application of “conservation of charge”.

Page 45: W = V 2 t/R  R = V 2 t/E 115  W .
Page 46: W = V 2 t/R  R = V 2 t/E 115  W .

Total Voltage = sum of all partial voltages on circuit. Application of conservation of energy.

2 V 4V

6V

Page 47: W = V 2 t/R  R = V 2 t/E 115  W .

What is the current in the other wire?

10 A

2A

16 A

?

4 A

Page 48: W = V 2 t/R  R = V 2 t/E 115  W .

What is the current in the unknown wire?

4A

12A

6 A?

2 A

Page 49: W = V 2 t/R  R = V 2 t/E 115  W .

Fuses – appliances are rated for the power they can safely dissipate.

That implies a certain current & voltage (power).

Fuses should be chosen to have a current rating a bit higher than the one for which the resistor is designed.

Page 50: W = V 2 t/R  R = V 2 t/E 115  W .

Should fuses and circuit breakers be connected in parallel or in series? Why?

Page 51: W = V 2 t/R  R = V 2 t/E 115  W .

If a 60 W bulb is connected to a 120 V source, the current is:

P = VI I = P/V.

I = 60 J/s = 0.5 C/s .5A.120 J/C

Fuse should be ~ .6-1 A.


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