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.

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 .

To represent components on circuit use symbols.

See ref tables

Sketching with Circuit Symbols

CircuitSymbols

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

symbols

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.

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?

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

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

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

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 …

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

1. 4.

2. 3.

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.

3. Given the circuit:

2-. 4-. 6-.

What is the equivalent resistance?

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

4. Given the circuit:

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

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

6V

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

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.

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

• 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.

Parallel Circuits

The current reaches a fork, can choose path.

What’s Different?

Through which resistor does more charge flow? Why?

Sketching: Make rungs or branches.

• 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

Resistance goes down as more branches are added:

1/Req = 1/R1 + 1/R2 + 1/R3 …

Where Req is the equivalent or total resistance.

*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!!

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.

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!

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.

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.

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?

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.

Hwk:Text pg 745 #2 - 5.

Sketch the circuits in #4, 5.

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.

Which equation is more useful for:

• Series?

• Parallel?

P = I2R

P = V2

R

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?

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?

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?

Use of Meters

Ammeters measure current so circuit current must flow through meter.

Connect meter in series to measure current.

Ideal ammeter has zero resistance.

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

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

Ideal voltmeter has infinite resistance.

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

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

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

2 V 4V

6V

What is the current in the other wire?

10 A

2A

16 A

?

4 A

What is the current in the unknown wire?

4A

12A

6 A?

2 A

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.

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

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.