Phy 213: General Physics IIIElectromotive Force & Internal ResistanceKirchoff’s Voltage & Current LawsSeries CircuitsParallel CircuitsAnalyzing Circuits 1 (using Kirchoff’s Laws)Analyzing Circuits 2 (using equivalent resistances)RC CircuitsCharging a CapacitorCapacitive Charging & Discharging (C=0.1F, Vmax=10V & R=10W)Capacitive Charging in RC circuit (Effects of increasing R on Vcap)Capacitive Charging (i vs t)Phy 213: General Physics IIIChapter 27: Electric CircuitsLecture NotesElectromotive Force & Internal Resistance1. The intrinsic potential difference associated with a power source is referred to as the “electromotive force” or emfa. In an ideal power source, the voltage across its terminals is its emfb. For a real power source, such as a battery, the emf is determined by the net electrochemical potential due to its internal redox reaction BUT the actual voltage across its terminals is slightly lower due to its internal resistance (Rint).2. Real power sources are limited in their ability to deliver power output, due to factors such as the maximal rate of the internal chemical reaction, the input power (in an AC plug-in DC power source), etc…a. For a battery, the rate of reaction is dependent on the conditioning and corrosion of electrodes and depletion of internal reactants. This results establishes an effective internal resistance, within the voltage source.3. As an electrochemical power source is utilized and is run down, the decline in performance output is reflected in the increased in internal resistancea. The output voltage will wane as more of the potential drops across Rint even though the emf remains constantRint+-+-emfVExample: A single loop circuit where R1=R2=R3=10Kirchoff’s Current Law (aka the Node Law):•The total current through any node is equal to zero:Kirchoff’s Voltage & Current LawsKirchoff’s Voltage Law (aka the Loop Law):•For any closed loop in a circuit, the total voltage around the loop is equal to zero:1 2 3 1 2 3Ni R R R R R R Ri=1RR loopV =V-V -V -V =0 V +V +V =3V =10VVV = 3.33V i = = 0.333A Rand��+-10VR1R3R2Nii=1V = 0�Nii=1i = 0�i1i3i2Series Circuits1. In series wiring, circuit elements (loads) are connected end to end2. The combined load or resistance (Req) in the series is3. Across each resistance, the potential difference (V) drops4. The current i that flows through R1 also flows through R2V = V1 + V2 = iR1 + iR2orV = i(R1 + R2)= iReqNeq 1 2 N ii=1R =R +R +...+R = R�VParallel Circuits1. Circuit elements (loads) are connected with ends attached2. The combined load or resistance (Rp) in the parallel is3. Across each resistance, the potential difference (V) is the same4. The total current drawn through the circuit is: i = i1 + i2or Ni=1eq 1 2 N i1 1 1 1 1= + +...+ = R R R R R�1 2 1 2 eqV V 1 1 Vi = + = V + = R R R R R� �� �� �1 21 2i iV = = R RAnalyzing Circuits 1 (using Kirchoff’s Laws)Consider the following 2 loop circuit, with 6 equal value resistors (R=R1=R2=R3=R4=R5=R6=10:1. What is the current & voltage for each R?a. Kirchoff’s Laws:Loop 1:Loop 2:Node:solving for i’s:1 2 3R R R1 1 2 2 1 3V +V +V =10Vi R +i R +i R =10V( )4 5 6 3R R R R3 4 5 6 2 2V +V +V =Vi R +R +R =i RR6R3+-10VR1R2R4R512i1i3i2i11 2 3i = i + i( ) ( )( ) ( )( ) ( )4 5 621 313 R R R112 3 R1 2 3 R Ri = A=0.091 A V =V =V = 10 0.091A =0.91V i =3i =0.273 A V = 10 0.273A =2.73V i =i +i =0.364 A V =V = 10 0.364A =3.64V � W� W� WAnalyzing Circuits 2 (using equivalent resistances)Consider the same circuit:2. What is the current & voltage for each R?a. Solve for Req1:b. Solve for Req2:c. Solve for Req:d. Use Req to get i1 & Req2 to get VR1& VR3e. Solve for the rest:R3+-10VR1R2Req12 eq1eq22 eq1R RR = = 7.5R +RWeq1 4 5 6R = R +R +R = 30WR3+-R1Req2+-Reqeq 1 eq2 3R =R +R +R =27.5W( ) ( )1 R1 R310Vi = =0.364A V =V = 0.364A 10V =3.64V27.5�WR2 eq22 3V =V =(10-7.27)V=2.73V2.73V 2.73Vi = =0.273A and i = =0.091A10 30�W WRC Circuits1. A circuit containing a capacitor and resistor(s) is called an RC circuit2. A resistor in series with a capacitor will limit the rate (not quantity) at which charge accumulates in the capacitor3. When V is constant across a capacitor ( = 0) no current will flow through this branch of the circuit since:dq dVi= =C = 0dt dtdVdtCRdq dqq = CV = CiR = RC = RCdtdt q�maxqtmaxq 0integrating both sidesdq 1 q 1 = dt ln = tq RC q RC� �� �� �� �� �t-RCmaxsolving for q q(t) = q e�4. When a fully charged capacitor is discharged, the rate of charge loss is limited by the voltage across it and is limited by on the resistance:Charging a CapacitorThe voltage equation around a loop with resistor and capacitor in series with a constant voltage source is given by:Re-arranging the equation leads to a 1st order linear non-homogeneous differential equation. This can be solved by applying the separation of variables technique:CVRcapR capV - V - V = 0q dq q VV - iR - = 0 +q dqwhere V = and i=C = C dt RC Rdt� ������maxmaxqtmaxmax max0 0t-RCmaxqdq V q q dq 1 = - = - = dtdt R RC RC RC q -q RCq -qdq 1 t = dt ln = q -q RC q RCq(t) = q 1 - e�� ��� �� �� ���� �� ��� �Capacitive Charging & Discharging(C=0.1F, Vmax=10V & R=10)-tRCmaxq(t) = q 1 - e� �� �� �-tRCmaxq(t) = q eCapacitive Charging in RC circuit(Effects of increasing R on Vcap)Charging Discharging0246810120 10 20time (s)V (volts)10 ohms30 ohms60 ohms0246810120 10 20time (s)V (volts)10 ohms30 ohms60 ohmsCapacitive Charging(i vs t)Charging DischargingWhat should these graphs should look