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Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2 , Slide 1

Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

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Page 1: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Electricity and MagnetismReview 2: Units 7-11

Mechanics Review 2 , Slide 1

Page 2: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

R3

V

R1 R2

C

S

Immediately after S is closed: what is VC, the voltage across C?

Example: RC Circuit

In this circuit, assume V, C, and Ri are known. C is initially uncharged and then switch S is closed.

What is the voltage across the capacitor after a very long time ?

VC = 0 what is I2, the current through R2?

At t = 0 the capacitor behaves like a wire. Solve using Kirchhoff’s Rules.

V

R1 R2

S

R3

Page 3: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

V

R1 R2

C

R3

S

Immediately after S is closed, what is I1, the current through R1 ?

V

R1 R2

S

R3

VC = 0

I1

Example: RC Circuit

In this circuit, assume V, C, and Ri are known. C is initially uncharged and then switch S is closed.

What is the voltage across the capacitor after a very long time ?

Page 4: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: RC Circuit

V

R1 R2

C

R3

S

After S has been closed “for a long time”, what is I2, the current towards C ?

V

R1

R3I2 = 0

VC

I

In this circuit, assume V, C, and Ri are known. C is initially uncharged and then switch S is closed.

What is the voltage across the capacitor after a very long time ?

After a long time the capacitor and R2 are are not connected to the circuit.

Page 5: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

V

R1 R2

C

R3

S

VC = V3 = IR3 = (V/(R1 + R3))R3

V

R1

R3VC

I I

In this circuit, assume V, C, and Ri are known. C is initially uncharged and then switch S is closed.

What is the voltage across the capacitor after a very long time ?

Example: RC Circuit

31

3

RR

RV

Page 6: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

In this circuit, assume V, C, and Ri are known. C initially uncharged and then switch S is closed for a very long time charging the capacitor. Then the switch is opened at t = 0.

Redraw the circuit with the switch open. Now it looks like a simple RC circuit.

What is tdisc, the discharging time constant?

What is the current on R3 as a function of time?

R2

C R3

Example RC Circuit

V

R1 R2

C

R3

S

Page 7: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Capacitors

Three capacitors are connected to a battery as shown. A) What is the equivalent capacitance?B) What is the total charge stored in the system?C) Find the charges on each capacitor.

V

Parallel:C23 = C2 + C3

Series: (1/C123) = (1/C23) +

(1/C1)Total Charge:

Q = C123 V

Charges on capacitors:Q1 = Q Q2 + Q3 = Q

V2 = V3

Page 8: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

+ -

+ -

+ -

Batteries are easy

V1R1

R2

In this circuit Vi and Ri are known.What are the currents I1 , I2 , I3?

R3

V2

V3

I1

I3

I2

Label and pick directions for each currentLabel the + and - side of each element

- +

+ -

- +

For resistors, the “upstream” side is +

Example: Kirchhoff’s Rules

Page 9: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

+ -

+ -

+ -

V1R1

R2

In this circuit Vi and Ri are known.What are the currents I1 , I2 , I3?

R3

V2

V3

I1

I3

I2

- +

+ -

- +

Example: Kirchhoff’s Rules

1. I2 = I1 + I3

2. - V1 + I1R1 - I3R3 + V3 = 0

3. - V3 + I3R3 + I2R2 + V2 = 0

4. - V2 - I2R2 - I1R1 + V1 = 0

We need 3 equations: Which 3 should we use?

Kirchhoff’s Rules give us the following 4 equations:

The node equation (1.) and any two loops.

Page 10: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

First determine E field produced by charged conductors:

Integrate E to find the potential difference V

Example: Calculating Capacitance

A solid cylindrical conductor of radius a length l and charge Q is coaxial with a thin cylindrical shell of radius b and charge –Q. Assume l is much larger than b.

Find the capacitance

Page 11: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

First determine E field produced by charged conductors:

Integrate E to find the potential difference V

Example: Calculating Capacitance

A spherical conducting shell of radius b and charge –Q is concentric with a smaller conducting sphere of radius a and charge Q. Find the capacitance.

Page 12: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Capacitors

In the circuit shown the switch SA is originally closed and the switch SB is open.

(a) What is the initial charge on each capacitor.

Then SA is opened and SB is closed.

(b) What is the final charge on each capacitor.

(c) Now SA is closed also. How much additional charge flows though SA?

Initial Charge:Q1i = C1 ΔV Q2i = Q3i = 0

Final Charge:Q1f = C1 ΔVf Q2f = Q3f

ΔVf = Q2f /C2 + Q3f /C3 Q1i = Q1f + Q2f

Both switches closed:Qtotal = Ctotal ΔV Ctotal = C1+1/(1/C2+1/C3)

Page 13: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Capacitor with Dielectric

An air-gap capacitor, having capacitance C0 and width x0 is connected to a battery of voltage V. A dielectric (k ) of width x0/4 is inserted into the gap as shown. What is Qf, the final charge on the capacitor?

What changes when the dielectric added?

VC0

x0 kV

x0/4V

QC

C and Q change,V stays the same.

Page 14: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Capacitor with Dielectric

Can consider capacitor to be two capacitances, C1 and C2, in parallel

C1 = 3/4C0What is C1 ? For parallel plate capacitor: C = e0A/d

=C1

C2k k

What is C2 ? C2 = 1/4 k C0

What is C ? C = C0 (3/4 + 1/4 k)

What is Q? Qf = VC0 (3/4 + 1/4 k)

V

QC

Page 15: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Circuits

Redraw the circuit using the equivalent resistor R24 = series combination of R2 and R4.

R2 and R4 are connected in series (R24) which is connected in

parallel with R3

V

R1 R2

R4

R3

In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, and R4 = 4W. What is V2, the voltage across R2?

V

R1

R3 R24

R24 = R2 + R4 = 2W + 4W = 6W

First combine resistors to find the total current:

Page 16: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Circuits (Without Kirchhoff’s Rules)

R3 and R24 are connected in parallel = R234

1/Req = 1/Ra + 1/Rb

In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, and R4 = 4W. What is V2, the voltage across R2?

V

R1

R3 R24

1/R234 = (1/3) + (1/6) = (3/6) W -1

= 2 W

V

R1

R234

R1 and R234 are in series. R1234 = 1 + 2 = 3 W

= I1234

V

R1234

Ohm’s Law

I1 = I1234 = V/R1234

= 6 Amps

Page 17: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

a

b

Example: Circuits

R234

Since R1 in series with R234

I234 = I1234 = I1 = 6 Amps

V234 = I234 R234 = 6 x 2 = 12 Volts

In the circuit shown: V = 18V, R1 = 1W, R2 = 2W, R3 = 3W, and R4 = 4W.

R24 = 6W R234 = 2 W I1234 = 6 A

What is V2, the voltage across R2?

= I1234

V

R1234

V

R1

R234I1 = I234

What is V234 (Vab)?

Page 18: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Circuits

V

R1

R24R3

I24 = I2 = 2Amps

V

R1 R2

R4

R3

I1234

Ohm’s Law

V2 = I2 R2

= 4 Volts

I24

I24 = V234 / R24 = 2Amps

Page 19: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Circuits

What is I3 ?

V

R1

R234

a

b

V

R1 R2

R4

R3 =V = 18V

R1 = 1W R2 = 2W R3 = 3W R4 = 4WR24 = 6W R234 = 2WV234= 12VV2 = 4VI1 = 6 AmpsI2 = 2 Amps

I1 I2

I3

I1 = I2 + I3

I3 = 4 A

What is P2 ? P2 = I2V2 = I22 R2 = 8

W

Page 20: Electricity and Magnetism Review 2: Units 7-11 Mechanics Review 2, Slide 1

Example: Kirchhoff’s Rules

Given the circuit below. Use Kirchhoff’s rules to find the currents I1, I2 and I3, and the charge Q on the capacitor. What is the voltage difference between points g and d? (Assume that the circuit has reached steady state currents)

I1 = 1.38 AmpsI2 = - 0.364 AmpsI3 = 1.02 AmpsQ = 66.0 μCVgd = 2.90 V