APC – UNIT 8DC Circuits

Whenever electric charges move, an electric current is said to exist. The current is the rate at which the charge flows through a certain cross-section A. We look at the charges flowing perpendicularly to a surface of area A

The time rate of the charge flow through A defines the current (=charges per time):

+ -

Current moving from + to – is called conventional current flow.

Atomic View of CurrentConsider a wire connected to a potential difference…

+ -E

Existence of E inside wire (conductor) does not contradict our previous results for E = 0 inside conductor. Why?

Current Density (j)

Current density is a vector field within a wire. The vector at each point points in the direction of the E-field

Drift VelocityCurrent was originally thought to be positive charge carriers (Franklin) and therefore that became conventional current flow. However, it is the free electrons (valence) that move but they encounter many collisions with atoms in the wire. (non-conventional)

In a time Δt, electrons travel a distance Δx = vd Δt.Volume of electrons in Δt pass through area A is given asV = A Δ x = A vd Δ tIf there are n free electrons per unit volume (n= N/V) where N = # of electrons then the total charge through area A in time Δt is given bydQ = (# of charges)x(charge per e-)

dQ = nV(-e) = -n A vd (dt) e

Minus means dirn of + current opposes dirn of vd

Also…

We can relate the current to the motion of the charges

Ohm’s Law

Materials that obey Ohm’s Law are said to be ohmic. Incandescent light bulbs are non-ohmic.

Conductivity

Within a wire of length L, I = jA and V = EL, substituting into Ohm’s Law we get

High resistivity produces less current density for same E-field

Conductivity is the reciprocal of resistivity.

+ -E

Vb Va

Electrical PowerConsider the simple circuit below. Imagine a positive quantity of charge moving around the circuit from point A through an ideal battery, through the resistor, and back to A again.

A

BAs charge moves from A to B through battery, its electrical energy increases by an amount QΔV while the chemical PE of battery decreases by that amount.

When the charge moves through the resistor, it loses EPE as it undergoes collisions with atoms in R and produces thermal energy.

A

BThe rate at which charge loses PE in resistor is given by

From this we get power lost in the resistor:

Series Circuit Characteristics

Parallel Circuit Characteristics

Short Circuit

Ammeter and VoltmeterAMMETERS have a very small resistance to limit their effect on introducing resistance into the circuit being measured. Connected in SERIES.

VOLTMETERS (V) have a very large resistance to reduce the amount of current drawn from the circuit being measured (short). Connected in PARALLEL.

Compound Circuita) V = 10V, find the potential difference across R4

b) If the wire before R2 is cut (inoperable) what happens to the total current?

d) RESET…If R2 is replaced by a wire what happens to the total current?

c) If wire is cut at C between R3 & R4 what happens to VA? VX?

RESET problem

C

If V = 45V, determine the power dissipated in R5.

Potentiometer or Variable ResistorDevice that allows for you to vary the resistance by changing the effective length of wire

symbol

EMF (electromotive force),εA ideal battery has no internal resistance (friction). However, a real battery has some internal resistance where there is a voltage drop within battery leaving less ΔV for external circuit.

The voltage available for external circuit is called the terminal voltage, Vab . The internal resistance is r. Therefore,

When battery is not connected or ideal

Different sized batteries (AAA vs D) have different amp-hour ratings. The larger the battery, the higher the amp-hour rating for the same V. Larger-sized batteries have more charge to supply

The battery capacity that battery manufacturers print on a battery is usually the product of 20 hours multiplied by the maximum constant current that a new battery can supply for 20 hours at 68 F° (20 C°), down to a predetermined terminal voltage per cell. A battery rated at 100 A·h will deliver 5 A over a 20 hour period at room temperature.

Series and Parallel EMFsBattery Charging

EMFs in series in the same direction: totalvoltage is the sum of the separate voltages…it is increased.

EMFs in series, opposite direction: totalvoltage is the difference, but the lower voltagebattery is charged.

EMFs in parallel are not used to increase voltage but to provide more energy when large currents are needed. Each cell only produces a fraction of total current so loss due to internal resistance is less than for single cell. Batteries will last longer.

When connecting in parallel you are doubling the capacity (amp hours) of the battery while maintaining the voltage of the individual batteries

Batteries MUST be the same, If not, there will be relatively large currents circulating from one battery through another, the higher-voltage batteries overpowering the lower-voltage batteries.

In this case, VR = 12V and if R=1, IT = 12A with each battery providing only 6A each.

Power delivered to Load (R)

When is the power delivered to the load a maximum when battery is NOT ideal?

Kirchoff’s Rules

1) Junction Rule ( Ij = 0) (conservation of charge)The sum of the currents entering any junction must equal the sum of the currents leaving thatjunction.

Circuits that are complex in that they cannot be reduced to series or parallel combinations require a different approach.

2) Loop Rule (Vj ) = 0 (conservation of energy)

The sum of the potential differences across all the elements around any closed circuit loop must be zero. (valid for any closed loop);

Kirchoff Example Calculate the current in each branch of the circuit.

If a voltmeter was connected between points c and f, what would be the reading (Vcf)? Vcf means Vc – Vf.

Ideal batteries ξ1 = 11.5 V and ξ2 = 4.00V, and the resistances are each 3.2Ω.

a) What is the size and direction of current i1? (Take upward to be positive.)

b) What is the size and direction of current i2? (Take upward to be positive.)

c) At what rate is energy being transferred at the 4.00V battery and is the battery supplying or absorbing energy?

ξ2

ξ1i1

i2

11.5V 4.0V

a) Focus on left hand loop

ΣV around loop = 0, therefore V across resistor must be 11.5V. Moving CW around loop yields…

i1i2

i1 = -11.5V/3.2Ω = -3.59A

Negative since following direction of ξ1

b) What is the size and direction of current i2? (Take upward to be positive.)

4.0V

3.59Ai2

Focus loop on center rectangle

Move around loop, CW starting at battery.

4V + 3.59(3.2) – i2 (3.2) – (i2/2)(3.2) – i2 (3.2) = 0

½ i2i2

i2 =1.94A, down

RC CIRCUITSOften RC circuits are used to control timing. Some examples include windshield wipers, strobe lights, and flashbulbs in a camera, some pacemakers.

ξ

A closer look at current during charging process

Applying Kirchoff’s Loop Rule to RC circuit (CW) tofind V & q as function of time (after closing switch):

Time Constant, τThere is a quantity referred to as the time constant of the RC circuit. This is the time required for the capacitor to reach 63% of its charge capacity and maximum voltage. It also represents the time needed for the current to drop to 37% of its original value.

It can be shown that after 1 time constant (RC), VC is 63%

of its maximum voltage, Vo.

Note: R needs to be in series in some way with C for there to be an effect on the time constant (explain later)

0.37 Imax

0.63 Vmax

1 time constant, RC

Discharging

ξ

After a very long time, the capacitor would be fully charged. If the switch was then moved to ‘b’…

i

ExampleBoth switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S1 is closed?

What is the current through the battery after switch 1 has beenclosed a long time?

a) Ib = 0 b) Ib = ε / (3R)

c) Ib = ε /(2R) d) Ib = ε / R

b) Ib = V/(3R)

a) Ib = 0

c) Ib = V/(2R)

d) Ib = V/R

2R

RC

S1 S2

ε

Both switches are initially open, and the capacitor is uncharged. What is the current through the battery just after switch S1 & S2 are closed?

After a long time what is the current through the battery?

a) Ib = 0 b) Ib = ε / (3R)

c) Ib = ε /(2R) d) Ib = ε / R

b) Ib = ε /(3R)

a) Ib = 0

c) Ib = ε /(2R)

d) Ib = ε /R

After a long time S1 is opened. What is the voltage across R and 2R after 2τ?

2R

RC

S1 S2

ε

ExampleS

R2 R1

C

Find VR2 & VR1 after S has been closed for 1τ. 12V

S

ExampleS

R2 R1

C

Find total current at this time (1 τ ) if R1 = 10Ω and R2 = 20Ω

12V

Each circuit below has a 1.0F capacitor charged to 100 Volts. When the switch is closed:

c) Which lights consumes more energy assuming we wait until both can’t be seen?

Example

a) Which system will be brightest?

b) Which lights will stay on longest?

18V

1MΩ

3MΩ 6MΩ

1µF

A capacitor is initially uncharged and then the switch is closed.

a) At t = 0, when switch is closed, find the current in each resistor.b) Find these currents after a long time later.

c) Find the voltage drop on the capacitor just after the switch is opened.

After a terribly long time, the switch is opened.

18V

1MΩ

3MΩ 6MΩ

1µF

18V

1MΩ

3MΩ 6MΩ

1µF

d) Find the charge on the capacitor 18s after the switch is opened.

e) How much energy is consumed by the 6MΩ during the discharging process?