Reading Quiz – Chapter 31

1. Name or describe Kirchhoff’s Laws.

2. Which of the following are ohmic materials?

a. Batteries f. Materials a-d

b. Wires g. Materials b and c

c. Resistors h. Materials a and b

d. Light bulb i. All of a-e

e. capacitors

Learning Objectives – Ch 31

• To develop a conceptual model of simple circuits.

• To understand series and parallel resistances.

• To apply Kirchhoff’s laws to circuit analysis.• To find the connection between current and

potential difference for a conductor.• To understand energy transfer and power

dissipation in circuits.• To understand RC circuits.

ε and ∆V• For an ideal battery

potential difference between the positive and negative terminals, ∆V, equals the chemical work done per unit charge.

∆V = Wch /q = ε• ε is the emf of the battery• Due to the internal

resistance of a real battery, ∆V is often slightly less than the emf.

Ch – 31 Circuits

Resistors and Ohm’s Law

From the relationshipJ = σE = E/ρ

where ρ, resistivity = 1/ σIt can be shown that:

I = ∆V/R where R, resistance = pL/A This is Ohm’s LawUnits of resistance are Ω

(ohms)

Ohmic and nonohmic materials

Ohm’s Law is only valid for ohmic materials – where the current is proportional to the potential difference

Question

• Calculate the resistance of a 1.0 mm-diameter, 3 cm-long carbon resistor. Table 30.1, p.944 gives resistivity values:

Answer: 1.34 Ω

• Calculate the resistance of a 1.0 mm-diameter, 3 cm-long carbon resistor. Table 30.1, p.944 gives resistivity values:

Ohmic materials: the “Ideal” wire

• Wires are metals with a very very small resistance (R<< 1 Ω)

• The ideal wire has a R=0 Ω, therefore ∆V = 0 V between the ends of an ideal wire.

Hmmm….

• Recall that Ohm’s Law is I = ∆V/R

• If the wire at the right is an ideal wire, will the value of the current be:

a. undefined and therefore 0

b. Huge

Ohmic materials: Resistors

• Resistors are conductors, albeit poor ones, that are used so the current in a circuit of ideal wires will not be “huge”. Typical Range of R: 10 to 106 Ω.

• A light bulb filament is a non-ohmic resistor.

Ohmic Materials - Insulators

• Insulators are materials such as glass, plastic or the space between two parts of a wire after you have removed a resistor or light bulb (open circuit). The ideal insulator has a resistance of infinity.

• Recall that Ohm’s Law is I = ∆V/R . Will an open circuit result in: 1. current of zero

2. potential difference of zero

3. both values going to zero

Kirchoff’s Loop Law

• The sum of all potential differences encountered while moving around closed path is zero

• This is a result of the conservation of energy for a conservative force

Kirchoff’s Junction Law

• The total current into a junction must equal the total current out of the junction

• Based on conservation of charge and current

Σ Iin = ΣIout

Basic Circuit for the Carbon Resisitor

• Calculate the current that leaves the positive terminal of the ideal battery (ε= 4V).

• Is this current greater than, less than or equal to the current at leaving the bottom end of the load?

Answers

• 2.99 A• Equal to: Current out

of the battery must equal current into it according to Kirchoff’s Junction Law;

Energy and Power

Power is that rate at which energy is transferred

Pbattery = rate at which the battery supplies energy to the charges

Pbattery = of energy transfer = dU/dt

Pbattery = (dq/dt)ε

Pbat = I ε

What happens to the energy?

• The chemical energy of the battery is converted to U, electrical potential energy: Echem U

• The resulting electric field causes the electrons to accelerate: UK

• Collisions in the lattice structure transfer the energy to the lattice as thermal energy: KEth

• Thermal energy is a dissapative energy (i.e. can’t be recovered like mechanical energy.

What happens to the energy?

• The rate at which energy is transferred from the current to the resistor is:

PR = dEth/dt

PR = (dq/dt)∆VR

PR = I ∆VR

• Since a resistor obeys Ohm’s Law:

PR = I2R = (∆VR)2/R

Numerical question

What is the magnitude and direction is the current in the 30 Ω resistor?

Draw a graph of the potential as a function of the “distance” traveled through the circuit (see ex 31.2 for distance

Answer

a. Current = 0.100 A

Power rating of a light bulb or household applicance

• Commercial and residential electrical systems are set up so that each individual appliance operates at a potential difference of 120 V.

• Power Rating or Wattage is the power that the appliance will dissipate at a potential difference of 120 V.

• Power consumption will differ if operated at any other potential difference (i.e. 220, such as is standard in Europe and many other countries).

Resistors in Series• Resistors aligned from end to end, with no junctions between them

are in series• The current I is the same through each of the resistor• The potential difference through each resistor is :

∆VR = (-)IRThe equivalent resistor is made by adding up all resistors in series in

the circuit.

Resistors in series question

• What is the Value of R?

Answer: R = 25Ω

Real Batteries

• Real batteries have an internal resistance, r, and therefore:

∆V = (ε – Ir)< ε

∆V = R/( R +r) ε#19 in workbook

Workbook Problem #14

A 60 W light bulb and a 100 W bulb are placed one after the other in a circuit. The battery’s emf is 120V. Which one glows more brightly? Assume the resistance of the bulbs to be ohmic.

Answer

• The 60 W bulb.

• Calculate the resistance of each bulb, knowing the power rating when it is operated alone at 120V. Which has the higher resistance?

• Calculate the equivalent resistance, and the current they both draw when in the series circuit

Resistors in Parallel

• Resistors that are connected at both ends are considered to be in parallel.

• Connecting resistors in this manner increases the amount of current since it increases the number of paths.

• The left end of R1 has the same potential as the left end of R2 and the same is true of the right end. Therefore the potential difference (∆V) across each resistor connected in parallel is equal.

Resistors in Parallel

• Kirchoff’s current law applies at the junctions. Ic = Id. Therefore the current through each of the branches depends on the resistance of the resistor

• The equivalent resistance:

Rcd = (1/R1 + 1/R2)-1

Parallel resistor question

What is the value of R?

Answer

• IR = .667A

• R = 12 Ω

Complex resistor problem

• Find the current through the circuit.• What are the values of potential at points a-f?

Complex resistor problem

• Answer: 1A• a:0V b:12V c:9V d: 6V e: 0V f: 0V

The circuit is grounded at point c

What is the current through the circuit?

What are the values of potential at points a-f?

The circuit is grounded at point c

What is the current through the circuit? – still 1A

What are the values of potential at points a-f?

a: -9V b: 3V c: 0V d: -3V e: - 9V, f: -9V

The Capacitance of a Capacitor (Ch. 30)

• 2 expressions for the electric field of a capacitor:

E = Q/(ε0 A)

relates electric field to surface charge density

E = ∆Vc/d

Electric field is the rate of change of the potential across plates

The Capacitance of a Capacitor (Ch. 30)

• Combining the 2 expressions:

Q = [(ε0 A)/d]∆Vc

• potential difference of the capacitor is proportional to the charge on the plates.

• The constant of proportionality is a geometric factor

The Capacitance of a Capacitor (Ch. 30)

• Q = [(ε0 A)/d]∆Vc

• Capacitance (C) = (ε0 A)/d. For a given charge, a capacitor with a larger capacitance will have a greater potential difference.

• The SI unit of capacitance is the farad:1 farad = 1 F = 1 Coulomb/Volt.• Typical capacitors have values in the

microfarad to picofarad range.

RC Circuit

• Consists of a capacitor and a resistor.

• Unlike a battery, a capacitor cannot provide a constant source of potential difference.

• This value is constantly changing as the charge leaves the plate.

• Current due to a discharging capacitor is finite and changes over time.

RC Circuit

• Use Kirchoff’s loop law to analyze the circuit, starting at the capacitor:∆Vc + ∆VR = 0 Q/C – IR = 0

• Since charge is being removed from the plates:

I = - dQ/dtQ/RC + dQ/dt =0

• This is a differential equation, but is relatively easy to solve by rearranging terms:

dQ/Q = -[1/(RC)]dt• And then by integrating….

RC Circuit• The initial conditions

were Q = Q0 at t=0 (switch closes)

Q

Qo

QdQ / t

dt0

= -1/RC

ln Q – ln Q0 = -t/RC

ln (Q/Q0) = - t/RC

RC Circuitln (Q/Q0) = - t/RCTo solve for Q, capacitor charge at time t, take the exponential of both sides and multiply by Q0:

Q = Q0e-t/RC

the quantity RC is called the time constant, τ:

Q = Q0e-t/ τ

I = dQ/dt = I0e-t/τ

Charging a capacitor

• After the switch is closed, the battery will move charge from the bottom plate to the top plate of the capacitor.

• As the charge on the capacitor increases, the current through the circuit will decrease.

• Current ceases when ∆Vc = ε.

Q = Qmax(1 – e-t/τ)

Capacitor Problem

• A 10 μF capacitor initially charged to 20 μC is discharged through a 1 kΩ resistor. How long does it take to reduce the capacitor’s charge to 10 μC?

Capacitor Problem

• A 10 μF capacitor initially charged to 20 μC is discharged through a 1 kΩ resistor. How long does it take to reduce the capacitor’s charge to 10 μC?

Answer: 10 μC = (20 μC) et/0.010 s

Take the natural logarithm of both sides:

ln (10 μC/20 μC) = - t/0.010 s

= 6.93 s