Time Level of concentration 5.00pm

Syllabus Reactive components: Inductors and Capacitors. Ohms law, resistors in series and in parallel. Power. Ideal and realistic Voltage and Current sources. Kirchoff’s Laws. Resistive networks. AC circuits.

Moving charges and electric currents Battery + – Electric current I 1 ampere = 1 A= 1 Coulomb per second = 1 C/s

Current is scaler I1I1 I2I2 I3I3 The arrows that we use in diagrams are indicative of the direction of the current flow. The arrows point in the direction in which positively charged particles would be forced to move by the electric field.

1 A 2 A 3 A 4 A i Find the direction and magnitude of current i in the circuit above.

Current Density

Resistance Pump Water current proportional to potential difference Water current inversely proportional to pipe resistance

Electrical resistance R Battery = Charge pump High potential Low potential Poential difference = V I If V= 1 V, I= 1 A, then R=1  Electrical resistance R

Electrical Power I V In time dt, charge dq moves through the resistor. (potential drop V). Thus energy drop is (dq V). To maintain the current, the battery must lift charge dq to the potential V, performing work To work done in unit time is the power of the battery:

emf devices and internal resistance A device, such as a battery, which can maintain a potential difference by pumping charges is called an emf device. emf devices: electric generator uses mechanical energy to pump charges. Solar cells

An ideal emf device : the potential difference between the terminals of an ideal emf device is always  A real emf device, has internal resistance, so that the potential difference between its two terminals depends on what load is connected.

a I R r  + – b R r  a a b i ir iR  Potential The voltage available for R:  - ir. The internal consumption: ir, proportional to i.

The loop rule: The sum of the changes in potential encountered in a complete traversal of any loop of a circuit must be zero. This is also known as Kirchhoff’s loop rule. For a move through a resistance in the direction of the current, the change in potential is –iR. For a move through an ideal emf device in the direction of the emf arrow, the potential change is .

a I R r  + – b Apply Kirchhoff’s loop rule to the above circuit: Solving for the current: The power delivered to the external load: P reaches maximum for R=r

Resistances in series i  + – R eq R1  + – R2 R3

Kirchhoff’s Junction rule The sum of the current entering any junction must be equal to the sum of the current leaving the junction. I1I1 I3I3 I2I2 I4I4

Resistances in parallel R1R1  + – R2R2 R3R3 i i1i1 i2i2 i3i3 i 2 +i 3 The potential difference across all three resistors is the same  From Ohm’s law:

 + – R eq i R1R1  + – R2R2 R3R3 i i1i1 i2i2 i3i3 i 2 +i 3 Therefore:

For two resistances in parallel:

Sample problem 1 i R1R1  + – R3R3  R2R2  1 =4.4 V,  2 =2.1 V, R 1 =2.3 , R 2 =5.5 , R  Find i. Solution: use Kirchhoff’s loop rule

R1R1 i i  + – R3R3  R2R2

Sample problem 2 Each resistance is 1 , find the equivalent resistance  + –  + – R eq =4/3   + –

Some practical circuits 2. The Ammeter and the voltmeter 1.The Wheastone bridge

The Wheastone bridge R1 and R2 are standard Resistors with resistance values known. Rs is a variable resistance, its resistance value can be adjusted by sliding the contact  + – R1R1 R2R2 RsRs RxRx R0R0 a b A When point a and b are at the same potential:

 + – R1R1 R2R2 RsRs RxRx R0R0 a b A I1I1 I2I2 Potential drop across R1: I 1 R 1 Potential drop across Rs I 2 Rs I 1 R 1 =I 2 R s For the same reason: I 1 R 2 =I 2 R x

The Ammeter and the voltmeter The instrument to measure electrical currents is called an ammeter. The instrument to measure electrical potential differences is called a voltmeter.  + – R I

 + – R I  + – R I A  + – R I A V Resistance of an Ammeter must be very small Resistance of a voltmeter must be very large

Current source i Current source is a device that delivers a specified current I regardless of the load R i VRVR Current source is a device that delivers a specified current I regardless of V R

+ -  i R1R1 R2R2 Find voltage V V i1i1 i2i2 Kirchhoff’s loop rule around the red loop

The organic battery: electric eel

 R 500 electroplaques per row 140 rows

Physicists’ electric eel Total emf per row: 5000 x 0.15 V= 750 V Total resistance per row: 5000 x 0.25  = 1250  The equivalent circuit per row:  row = 750V R row =1250 

The equivalent circuit of many rows:  row = 750V R row =1250  R eq =?  eq = ?V

the electric eel stuns Why doesn’t the eel get itself stunned? Because the electric eel is FAT. When it comes to stunning, the current density is important, not just the current.

The electric eel is not an eel!!! Neotropical knifefishes Weight: 25 kg. 500 W electric power does not come from nothing.

Voltage divider R1R1 R2R2 V in V out

Network analysis Superposition theorem Eliminate all but one source at a time.

Take away B2, and calculate currents from B1 only R2//R3:I1= 6A, I2=2A, I3=4A.

Take away B1 R1//R2:I3= 3A, I2=2A, I1=1A. Now add algebraically: Current through R1:5A left to right; Current through R2: 4A down. Current through R3:3A

R1R1  + – R2R2 R1R1   R1R1 R1R1

R1R1  + – R2R2 R1R1   R1R1 R1R1 i1i1 i2i2 i3i3