Thursday, April 15, 1999
Announcements:
Lab Quiz next week on Ampere's Law. The lab cannot be purchased on the bookstore, but it can be downloaded from http://www.courses.psu.edu/phys/phys202_nxs16/sp99/

Lecture notes: Recall from Tuesday's lecture that an RL circuit can be compared to an Rc circuit.

• Note the similarity of an RL circuit with that of charging a capacitor.
• 
• At t = 0 the switch is closed.
• x - iR - q/C = 0            x - R dq/dt - q/C =0
• q = Cx( 1 - e^-t/t)

• x = - L (di/dt)
• Treat this as another emf, in the opposite direction of the emf due to the battery
• From Kirchhoff's Loop rule
x - iR - L di/dt = 0
• Note that after a long time, t,  di/dt= 0, so i = x/R
• We can guess then, that i = x/R( 1- e^-t/tL)
• What is tL ?
• After solving for the two equations, x - iR - L di/dt = 0 and i = x/R( 1- e^-t/tL), it can be determined that tL = L/R
• Therefore, in an RL circuit,
1. 1.)   i = x/R( 1- e^-t/tL), where tL = L/R
2. 2.) The voltage across R is iR...... V_R = x( 1- e^-t/tL)
3. 3.) The voltage across L is V_L = x_L = -L (di/dt) = -l (x/R)(1/tL)e^-t/tL = -xe^-t/tL

• There is an established DC current in the circuit. i_o = x/R

• At t=0, S2 is closed and S1 is opened (The battery is removed)
• Kirchhoff's Loop... -iR -L di/dt = 0
• iR + L di/dt = 0
• L di/dt = iR
• di/i = - R/L dt
• I(integral) di/i = -R/L I (integral)dt
• ln (i/i_o) = -Rt/L
• i/i_o = e^-Rt/L
• i = i_oe^-Rt/L
• Finally, i = i_oe^-t/tL , where tL = L/R
• Potential across R, V_R = iR
• V_R = iRe^-t/tL
• 
• Potential across L, x_L = -L di/dt
• di/dt = i_o (- 1/t_L) e^-t/tL = (i_oR/L)e^-t/tL
• So  x_L = -L di/dt = (i_oR/L)e^-t/tL
• Potential across L also decays exponentially

• The in-class demonstration proved that it is possible to generate a back emf that is larger than the emf supplied by the battery.

• x - iR - L di/dt = 0, so x = iR + L di/dt
• xi = i²R + Li di/dt, where xi is the rate the battery supplies energy, i²R is the rate the resistor dissipates energy
• Therefore, Li di/dt is the rate energy is stored in the coil.
• dU/dt = Li di/dt
• U = I(integral from 0 to i) Li di = (1/2) Li²
• For a solenoid, we have L = m_o n² Al, where A is the cross sectional area and l is the length
• Therefore, U= (1/2)(m_o n² Al)i² = (1/2m_o)(m_o ²n²i²) Al = (1/2m_o)B²Al
• U is magnetic is nature!
• u_B = U_B/ volume = (1/2m_o)B² = energy per unit volume

• It is an ideal circuit (R=0).The capacitor is charged up to Q =CV
• At t=0, the switch is closed. The charges in C will oscillate, i.e. the positive charges will move from one electrode to the other moving through L ( as a current).
• U_E = Energy contained in electric field of the capacitor = q²/2C
• U_Emax = Q²/2C
• U_B = Energy contained in the magnetic field of the inductor = (1/2) Li²
• U_Bmax = (1/2) LI²
• When the capacitor is fully charged U_E is at a maximum; there is no current and therefore U_B =0
• When the capacitor is fully discharged U_E = 0; Current is at a maximum and U_B is at a maximum
• At all times: U = U_E + U_B