Uses of Alternating Current
- over 99% of electrical energy used worldwide is produced by generators in the form of alternating current
- can be used to transport electricity at very high voltages (~400kV UK national grid) and therefore low currents reducing energy loss by heating
- these high voltages are then transformed to lower voltages with almost no energy loss for home usage (UK 240V)
AC Generator
-consists of a coil of area A with N turns rotating in a uniform magnetic field B
-the magentic flux Φb is given by:
Φb = N ∫ |B . ^n dA => Φb = NBA cosθ
-let the coil rotate with constant angular speed ω:
θ = ωt + 𝛿 , where delta is an arbitrary phase angle
-we can calculate emf generated across the terminals using Faraday’s Law:
ε = - dΦb/dt = - dBNAcos(ωt + 𝛿)/dt
= NBAωsin(ωt + 𝛿) = εmax * sin(ωt + 𝛿)
Ohm’s Law
V = IR
Current Flow
Description
- current in a conductor is driven by an electric field inside the conductor
- the electric field exerts a force on the charges within the material and the free charges then move and carry the current I
- as there is current flowing, the conductor is not in electrostatic equilibrium
- as there is an electric field across the conductor there is a change of electric potential ΔV across the material (written as V for neatness)
Resistors
Description
- conductors with resistance R
- treated as perfect ohmic materials
Capacitors
Solid Spherical Conductor - Capacitance
-by Gauss’s law, electric potential:
V = Q / 4πεoR
-rearrange for capacitance:
Q/V = 4πεoR = C
Capacitors
Description
- C is the capacitance and it is a measure of the capacity to store charge for a given potential difference
- units are Farads (F)
- a capacitor is a system of two conductors carrying equal but opposite charge Q
Capacitors
Parallel Plate Capacitors - Capacitance
-using Gauss’s Law:
V = Ed = σ/εo * d = Qd/Aεo
Q/V = Aεo/d = C
Capacitors
Parallel Plate Capacitors - Energy Stored
-work, W, has to be done to charge a capacitor and this goes into increasing the potential energy, U, of the capacitor
-consider a parallel plate capacitor with stored charge q and potential difference V across the plates
-move a small amount of charge dq from the negative plate to the positive plate:
dW = dU = Vdq
-since q = CV:
dU = q/C dq
-integrate
U = ∫ dU = ∫ q/C dq = 1/2 * Q²/C = 1/2 QV = 1/2 CV²
Where is the potential energy of a capacitor stored?
- when a capacitor is charged, an electric field is created between the two plates
- electromagnetic waves are just electric and magnetic fields and they transport energy
- this suggests that the energy in a capacitor is stored in its electric field
Capacitors
Parallel Plate Capacitors - Energy Desity
-starting from potential energy U: U = 1/2 CV² -sub in for C = εoA/d : U = 1/2 εoA/d V² -sub in for V=Ed : U = 1/2 εoA/d (Ed)² -for energy density, divide by volume which is Ad between the two plates ue = 1/2 εoA/d (Ed)² 1/Ad -simplify ue = 1/2 εo E²
Capacitors
Solid Spherical Conductor - Energy Stored
-start with electric field:
E = Q / 4πεor²
-from this electric potential is given by:
ΔV = - ∫ |E . d|l
-calculate the work done taking a small charge dq from infinity and adding it to the sphere which is currently at charge q:
dW = dU = dq ΔV = dq * q/4πεoR
-integrate between final total charge Q and 0
U = ∫ dq * q/4πεoR = 1/2 Q²/4πεoR
Capacitors
Solid Spherical Conductor - Energy Density
-take the expression for energy density for a parallel plate capacitor:
u = 1/2 εo E²
-consider a shell of thickness dr, a distance r from the centre of a solid sphere, the volume dV is given by:
dV = 4πr²dr
-using u for a parallel plate capacitor, dU stored in that volume is;
dU = u dV = u 4πr²dr = 1/2 εo E² * 4πr²dr
-sub in E = Q / 4πεor² and simplify:
dU = 1/2 εo (Q/4πεor²)² * 4πr²dr
dU = 1/2 Q²/4πεor² dr
-integrate over all space, i.e. from infinity to the edge of the sphere at r=R
U = ∫ 1/2 Q²/4πεor² dr = 1/2 Q²/4πεoR
-this matches the direct derivation of potential energy of a solid conducting sphere, therefore the energy density of electric field equation is general
Transient Current
Definition
- a current that changes over time before a steady state is reached is termed a transient current
- e.g. a transient flows after a switch that completes a circuit is closed
RC Circuit
Description
-consider a capacitor with capacitance C in a circuit with a resistor of resistance R, a battery with emf ε and a switch S
Solenoid - Self Inductance
-consider a solenoid of n turns per unit length carrying current I
-the current produces a magnetic field B inside the coil given by Ampere’s Law:
B = μo n I = μo (N/l) I
-magnetic flux through the coil:
Φ = NBA = N( μo (N/l) I )A
-multiply by length l to simplify;
Φ = N( μo (N/l²)AI l = (μon²lA)I
-the flux is proportional to current I, so we can write
Φ = LI
-where L = μon²lA is the self inductance of the coil
Self Inductance
L = Φ/I
- a constant of proportionality, where L = μon²lA
- units are the henry (H) where H=Tm²/A
Inductor
Description
-in a circuit we can add a component with inductance L. this component is called an inductor
-Faraday’s Law gives
ε = - dΦ/dt = -dLI/dt = -L dI/dt
-when the current through in an inductor changes the magnetic flux through the inductor changes which induces an emf in the inductor that opposes the change (Lenz’s Law)
Mutual Inductance
-when two circuits are close to each other, the magnetic flux through one circuit depends on the current in both circuits and we can define a mutual inductance M such that
Φb,1 = L1I1 + MI2
RL Circuit
Description
- resistor of resistance R and inductor of inductance L connected to a battery ε
- at time t=0 the switch is closed, on closing a current I will start to flow
- if I and dI/dt are positive then current increases in the direction of I
- a changing current in the inductor will induce an emf of magnitude L dI/dt that acts to oppose the change
- hence a positive dI/dt will induce an emf that tries o drive a current in the negative sense, so there will be a voltage drop across the inductor
- there is also a potential drop across the resistor of -IR:
RL Circuit
Power Equation
-starting with Kirchoff's Loop rule: ε - IR - L dI/dt = 0 -multiply each term by I and rearrange: Iε = I²R + IL dI/dt power output of battery = power dissipated as heat in the resistor + power into inductor
Inductor - Work Done
-to calculate the total work done W in setting up a current I flowing in the conductor start with power:
P = IL dI/dt
-since P=dW/dt, rearrange for dW:
dW = I (LdI)
-integrate between final current I and 0 for total work done
W = 1/2 LI²
-this energy must be stored in the magnetic field generated by the inductor
Inductor - Energy Density
-work done equals potential energy stored
U = W = 1/2 LI²
-sub in L = μon²lA :
U = 1/2 μon²lA I²
-since B = μo n I, sub in I = B/nμo
U = 1/2 μon²lA (B/nμo)²
-divide by the volume within a solenoid, V=Al to get energy density
um = U/lA = 1/2 μon² (B/nμo)² = 1/2 1/μo B²
Is energy stored in an inductor?
-it is easy to see that energy is stored in a capacitor since you can connect a charged capacitor to a circuit and it will power a current
-imagine a RL circuit with a switch positioned such that it is possible to bypass the battery
-move the switch so that it disconnects the battery from the inductor
ε - IR - L dI/dt = 0 becomes 0 - IR - L dI/dt = 0
-multiply by I and rearrange:
-LI dI/dt = I²R
power lost by inductor = power used to heat resistor
-this is true for all t so every joule lost by the inductor goes to doing something, in this case, heating the resistor
-without the inductor we could not have retrieved this energy and the difference between the inductor and the resistor is the magnetic field
