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Phy 213: General Physics III Chapter 27: Electric Circuits Lecture Notes Electromotive Force & Internal Resistance 1. The intrinsic potential difference associated with a power source is referred to as the “electromotive force” or emf a. b. 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. 3. 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. As an electrochemical power source is utilized and is run down, the decline in performance output is reflected in the increased in internal resistance a. + In an ideal power source, the voltage across its terminals is its emf 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). The output voltage will wane as more of the potential drops across Rint even though the emf remains constant Rint V + emf - - Kirchoff’s Voltage & Current Laws Kirchoff’s Voltage Law (aka the Loop Law): • For any closed loop in a circuit, the total voltage around the loop is equal to zero: N V i=1 i + 10V =0 V i=1 i R3 =V-VR1 -VR2 -VR3 =0 VR1 +VR2 +VR3 =3VR =10V VR = 0.333A R Kirchoff’s Current Law (aka the Node Law): • The total current through any node is equal to zero: VR = 3.33V and iloop= i2 i1 i3 R2 - Example: A single loop circuit where R1=R2=R3=10W N R1 N i i=1 i =0 Series Circuits 1. In series wiring, circuit elements (loads) are connected end to end 2. The combined load or resistance (Req) in the series is R eq=R1+R 2+...+R N= N R i=1 i 3. Across each resistance, the potential difference (V) drops V 4. The current i that flows through R1 also flows through R2 V = V1 + V2 = iR1 + iR2 or V = i(R1 + R2)= iReq Parallel Circuits 1. Circuit elements (loads) are connected with ends attached 2. The combined load or resistance (Rp) in the parallel is 1 1 1 1 = + +...+ = R eq R1 R 2 RN 1 i=1 R i N 3. Across each resistance, the potential difference (V) is the same i1 i2 V= = R1 R2 4. The total current drawn through the circuit is: i = i1 + i2 or 1 V V 1 V i= R1 + R2 = V + = R1 R2 R eq Analyzing Circuits 1 (using Kirchoff’s Laws) Consider the following 2 loop circuit, with 6 equal value resistors (R=R1=R2=R3=R4=R5=R6=10W): 1. What is the current & voltage for each R? a. Kirchoff’s Laws: Loop 1: VR +VR +VR =10V 1 2 Loop 2: VR +VR +VR =VR 5 6 10V + - 1 i1= i2+ i3 solving for i’s: i3 R2 i2 2 R5 3 i3 R 4+R 5+R 6 )=i2R 2 Node: R4 i1 3 i1R1+i2R2+i1R3=10V 4 R1 i1 R3 R6 1 i3= 11 A=0.091 A VR 4 =VR5 =VR6 =10W )0.091A )=0.91V i2=3i3=0.273 A VR2 =10W )0.273A )=2.73V i1=i2+i3=0.364 A VR1 =VR3 =10W )0.364A )=3.64V Analyzing Circuits 2 (using equivalent resistances) Consider the same circuit: 2. What is the current & voltage for each R? a. Solve for Req1: R eq1= R 4+R5+R 6= 30W b. Solve for Req2: R eq2= R 2R eq1 R 2+R eq1 10V = 7.5W R1 R1 + - R2 R3 Req1 + - Req2 R3 c. Solve for Req: R eq=R1+R eq2+R 3=27.5W d. Use Req to get i1 & Req2 to get VR1& VR3 i1= 10V =0.364A VR1=VR3=0.364A )10V )=3.64V 27.5W e. Solve for the rest: VR2=Veq2=(10-7.27)V=2.73V i2= 2.73V 2.73V =0.273A and i3= =0.091A 10W 30W + - Req RC Circuits 1. 2. 3. A circuit containing a capacitor and resistor(s) is called an RC circuit A resistor in series with a capacitor will limit the rate (not quantity) at which charge accumulates in the capacitor dV When V is constant across a capacitor dt ( = 0) no current will flow through this branch of the circuit since: dq dV i= 4. dt =C dt R C =0 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: dq dq q = CV = CiR = RC = RCdt dt q q t q dq 1 1 integrating both sides = dt ln = t q RC 0 RC qmax qmax solving for q - q(t) = qmaxe t RC Charging a Capacitor The voltage equation around a loop with resistor and capacitor in series with a constant voltage source is given by: R V - VR - Vcap = 0 V C q dq q V =0 + = C dt RC R q dq where V = and i= cap C dt V - iR - Re-arranging the equation leads to a 1st order linear nonhomogeneous differential equation. This can be solved by applying the separation of variables technique: qmax q dq V q dq 1 = = = dt dt R RC RC RC qmax -q RC t - t qmax -q t dq 1 0 qmax -q = RC 0 dt ln qmax = RC q(t) = qmax 1 - e RC q Capacitive Charging & Discharging (C=0.1F, Vmax=10V & R=10W) -t RC q(t) = qmax 1 - e 1.2 q (C) 1 0.8 q (charge) 0.6 q (discharge) 0.4 0.2 q(t) = qmaxe 0 0 5 10 time (s) 15 20 -t RC Capacitive Charging in RC circuit (Effects of increasing R on Vcap) Charging Discharging 12 12 10 8 6 10 ohms 30 ohms 60 ohms 4 2 0 V (volts) V (volts) 10 10 ohms 30 ohms 60 ohms 8 6 4 2 0 0 10 time (s) 20 0 10 time (s) 20 Capacitive Charging (i vs t) Charging Discharging What should these graphs should look like?