Reading Quiz – Chapter 31 • Name or describe Kirchhoff’s Laws. • Which of the following are ohmic materials? • Batteries f. Materials a-d • Wires g. Materials b and c • Resistors h. Materials a and b • Light bulb i. All of a-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.
Resistors and Ohm’s Law From the relationship J = σE = E/ρ where ρ, resistivity = 1/ σ It can be shown that: I = ∆V/R where R, resistance = pL/A This is Ohm’s Law Units 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: • undefined and therefore 0 • 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 106Ω. • 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: • current of zero • potential difference of zero • 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 = (-)IR The equivalent resistor is made by adding up all resistors in series in the circuit.
Resistors in series question • What is the Value of R?
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/dt Q/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) = -1/RC ln Q – ln Q0 = -t/RC ln (Q/Q0) = - t/RC
RC Circuit ln (Q/Q0) = - t/RC To 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