PHY 2049 Lecture Notes Chapter 28: Page 1 of 9 Electromotive Force E

I I +

Emf - DV V

Emf + DV = 0

Electromotive Force: The electromotive force EMF of a source of electric potential energy is defined as the amount of electric energy per Coulomb of positive charge as the charge passes through the source from low potential to high potential.

Emf = E = dU/dq (The units for EMF is Volts)

I + Single-loop circuit with a resistor: e R -

· Power Delivered by Emf (P = E I): dW dq dW = edq P = = e = eI dt dt · Power Dissipated in Resistor (P = I2R) · Combining the two:

E - IR = 0 and I = E /R (Kirchhoff's Rule)

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 2 of 9 Delivering power to a load: high voltage lines

I R +

Emf - Useful Load (motor, light bulb, etc)

· Power Delivered by Emf: P0 = E I

2 2 · Power Dissipated in wire resistance: P = I R= (P0/E ) R

2 · Losses: P/P0 = P0R/E

· Delivering power at high voltage results in lower losses

· Example:

Emf=120 V Power= 100 kW (this building) Line ~ 2×10-4 Ohm/m (1 cm ´ 1 cm copper) Losses <10%

Maximum length: L<0.1×1202 /105/0.0002 = 50 m

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 3 of 9 Circuit Rules

Loop Rule: Switch The algebraic sum of the changes in potential encountered in a complete traversal of any loop of a + circuit must be zero. C R Q - - IR = 0 . C

Junction Rule: The sum of the currents entering any junction must be equal the sum of the currents leaving that junction.

I = I å i å i I in out

DV=-IR Resistor: If you move across a resistor in the direction of the current flow then the potential change is DVR = - IR.

I Capacitor: If you move across a capacitor from minus to plus then the Q + potential change is

- DV = Q/C, DV=Q/C C and the current leaving the capacitor is I = -dQ/dt.

Inductor (Chapter 31): I If you move across an inductor in the direction of the current flow then the potential change is L DVL=-LdI/dt DVL = - L dI/dt.

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 4 of 9 Examples of multi-loop circuits

I I1 2

R1 R2 + + I1+I2 E1 - - E2 R3

To find currents through all resistors:

· Simplify: replace elements connected in series and parallel with corresponding equivalent elements

· Denote all “independent” loops

· Assign currents and their directions (directions can be picked arbitraryà solution will be negative if you picked wrong direction, that all)

· Apply the rule of current conservation at all junctions

· Go along each of the loops and add up all DV’s encountered; Sum of DV’s for a complete loop should be zero

First Loop : E1 - I 1 R1 - (I 1 + I 2 )R 3 = 0

Second Loop : E 2 - I 2 R 2 - (I 1 + I 2 )R 3 = 0

· Solve the system of equations

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 5 of 9 Voltmeter

Voltmeter: - device to measure difference of potentials between two points - connects in parallel (spy device) - should have large internal resistance

I R1 + Emf - DV=? RV R2

Error in the measurement: · Before connection: V0=ER2/(R1+R2) · After connection:

RV R 2 R 2 R 2 R 2 ® R 2¢ = = » R 2 (1 - d ) RV + R 2 1 + R 2 / RV 1 + d R ¢ R (1 - d ) æ R R ö 2 2 ç 1 2 ÷ V meas = E » E = ... = V 0 ç1 - ÷ R1 + R 2¢ R1 + R 2 (1 - d ) è R1 + R 2 RV ø

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 6 of 9 Ammeter

Ammeter: - device to measure current - a circuit is to be cut and the ammeter is inserted - should have small internal resistance

I=? R1 +

Emf -

R 2

RA

Error in the measurement: · Before connection: I0=E/(R1+R2) · After connection:

E E I 0 æ R A ö I = = = » I 0 (1 - d ) = I 0 ç1 - ÷ R + R + R æ R ö 1 + d ç R + R ÷ 1 2 A ç A ÷ è 1 2 ø ( R1 + R 2 )ç1 + ÷ è R1 + R 2 ø

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 7 of 9 Using Voltmeter and Ammeter to measure R

One has a battery, voltmeter, ammeter, and a resistor of unknown value. How should one connect the battery, resistor, voltmeter, and ammeter to evaluate R2 and to minimize error of the measurement?

Expected values of Emf (~9 V), R2 (~10 kOhm), RV (~100 MOhm), and RA (~10 Ohm) are known with about 10% uncertainty.

Homework problem…

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 8 of 9

RC-circuit: charging up a capacitor

After the switch is closed the current is entering the Switch capacitor so that I = dQ/dt, where Q is the charge on the + capacitor and summing all the potential changes in C going around the loop gives R Q - - IR = 0 C , where I(t) and Q(t) are a function of time.

If the switch is closed at t=0 then Q(0)=0 and Q dQ + R = 0 C dt , which can be written in the form dQ 1 = - Q dt t , where I have define t=RC.

Dividing by (Q-eC) and multiplying by dt and integrating gives t Q dQ 1 æ Q ö t lnç ÷ = - ò = -ò dt , which implies . Q0 è Q ø t Q 0 t 0

Solving for Q(t) gives Charging a Capacitor 1.50 1.25 Q(t) = Q e-t /t . 0 1.00

Q(t) 0.75

0.50

0.25 The curent is given by 0.00 I(t)=dQ/dt which yields 0 1 2 3 4 Q Time I (t) = 0 e-t /t RC . The quantity t=RC is call the time constant and has dimensions of time.

A. Korytov PHY 2049 Lecture Notes Chapter 28: Page 9 of 9 RC-circuit: discharging a capacitor

After the switch is closed the current is leaving the capacitor Switch so that I = -dQ/dt, where Q is the charge on the capacitor and + summing all the potential changes in going around the loop C gives R Q - - IR = 0 C , where I(t) and Q(t) are a function of time.

If the switch is closed at t=0 then Q(0)=Q0 and Q dQ + R = 0 C dt , which can be written in the form dQ 1 = - Q dt t , where I have defined t=RC.

Dividing by Q and multiplying by dt and integrating gives t Q dQ 1 æ Q ö t ò = -ò dt , which implies lnç ÷ = - . Q0 Q 0 t è Q0 ø t

Solving for Q(t) gives Discharging a Capacitor 1.00 -t /t Q(t) = Q0e . 0.75 Q(t) 0.50

0.25

The current is given by 0.00 I(t)=-dQ/dt which yields 0 1 2 3 4 Time Q I (t) = 0 e-t /t RC . The quantity t=RC is call the "time constant" and has dimensions of time.

A. Korytov