PHYS-2020: General Physics IICourse Lecture NotesSection IVDr. Donald G. LuttermoserEast Tennessee State UniversityEdition 3.3AbstractThese class notes are designed for use of the instructor and students of the course PHYS-2020:General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. Thesenotes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille.IV. Direct Current CircuitsA. Sources o f emf.1. Direct current means that the current travels only in one direc-tion in a circuit =⇒ DC for short.2. The source that maintains the constant current in a closed circuitis called a source of “emf .”a) One can think of such a source as a “charge pump” =⇒forces electrons to move in a d irection opposite the E-field,hence current, inside a source.b) The emf, E, of a source is the work done per unit charge=⇒ measu red in volts.c) A battery is a good example of a source of emf.3. Since batter ies have internal resistance, r, the potential differenceacross the terminals of a battery are∆V = E − Ir . (IV-1)a) E is equal to the terminal voltage when the current is zero.b) From Ohm’s law, the potential difference across the exter-nal resistance R, called the load resistance, is ∆V = IR,soIR = E − IrorE = IR + Ir = I(R + r) . (IV-2)c) If r  R, then E ≈ IR (as we have been assuming todate).IV–1IV–2 PHYS-2020: General Physics I IB. Resistors in Series.1. The current through all resistors in a series circuit is the same.I = I1= I2= · · · (IV-3)R1I1∆V1R2I2∆V2−+∆VII2. The reduced circu it above isReqIeq−+∆VIIwhere∆V = IeqReq= IReq∆V = ∆V1+ ∆V2= I1R1+ I2R2= IR1+ IR2IReq= IR1+ IR2.Donald G. Luttermoser, ETSU IV–33. As such, we can express the equivalent r esistance asReq= R1+ R2(IV-4)for two series resistors or more generally (e.g., for more than 2resistors in series) asReq=NXi=1Ri,series circuits(IV-5)where N is the total number of resistors.4. Once the equivalent resistance is found, one can then calculate Ifrom ∆V using Ohm’s law.C. Resisto rs in Parallel.1. When resistors are connected in parallel, the potent ial differen ceacross them are the same:∆V1 = ∆VR1I1∆V2 = ∆VR2I2−+∆VII⇓IV–4 PHYS-2020: General Physics I IReqIeq−+∆VII∆V = ∆V1= ∆V2. (IV-6)2. The current entering a parallel circuit is equ al to the sum of allcurrents t raveling through each resistor:I = I1+ I2(IV-7)in our 2 resistor diagram above.3. For the reduced circuit above:I =∆VReq, I1=∆V1R1, I2=∆V2R2,so∆VReq=∆V1R1+∆V2R2=∆VR1+∆VR2or1Req=1R1+1R2(IV-8)for two parallel resistors. If more that 2 r esistors are in a parallelconfiguration, we use1Req=NXi=11Ri.parallel circuits(IV-9)Donald G. Luttermoser, ETSU IV–54. Once the equivalent resistance is found, this in fo can be used todeduce either th e voltages or the individual cur rents across theresistors.Example IV–1. Consider the circuit shown below, where R1=3.00 Ω, R2= 10.0 Ω, R3= 5.00 Ω, R4= 4.00 Ω, and R5= 3.00Ω. (a) Find the equiv alent resistance of this circuit. ( b) If the totalpower supplied to the circuit is 4.00 W, find the emf of the battery.−+ER1R2R3R4R5Solution (a):We have to reduce this circuit in steps. But which do we reducefirst? Typically, one should look f or those resistors that are inthe most compact configuration and start there. For our circuitabove, resistors R2and R3have t he most compact confi gura-tion. Since these are in parallel, we use Eq . (IV-9) an d call theequivalent resistance Raas shown in the next figu re.1Ra=1R2+1R3=110.0 Ω+15.00 Ω1Ra=110.0 Ω+210.0 Ω=310.0 ΩRa= 3.33 Ω .IV–6 PHYS-2020: General Physics I I−+ER1RaR4R5We see fr om the diagram above that we now need to reduce the2 series resistors Raand R4with Eq. (IV-5) givingRb= Ra+ R4= 3.33 Ω + 4.00 Ω = 7.33 Ω .−+ER1RbR5The diagram below show the next stages on rou te to the finalequivalent resistor: Reduce the parallel Rband R5resistors fol-lowed the series Rcand R1resistors as shown in the next diagram.Donald G. Luttermoser, ETSU IV–7−+ER1Rc−+EReq=⇒Using the previous three circuits as a guide, we have1Rc=1Rb+1R5=17.33 Ω+13.00 Ω= 0.4697 Ω−1Rc= 2.13 ΩReq= R1+ Rc= 3.00 Ω + 2.13 Ω = 5.13 Ω .Solution (b):Since th e power is given to us, we can use Eq. (III-17) to de-termine the emf of the battery. If we assume that the intern alresistance of the battery is negligible to the load resistance, wehave ∆V = E, soP =(∆V )2Req∆V =qP · Req=q(4.00 W)(5.13 Ω)= 4.53 V .IV–8 PHYS-2020: General Physics I ID. Kirchhoff’s Rules and Complex DC Circuits.1. For complex DC circuits, it is not always possible to reduce all ofthe individual elements of the circuit into a single reduced circuit.2. In such cases, one should use Kirchhoff’s rules to figure outthe currents and potent ial differences across each element in thecircuit.a) The sum of the current entering any junction must equalthe sum of the cur rents leaving the junction =⇒ the junc-tion rule (see Figure 18.12 in your textbook).b) The sum of th e potential differences across all of the ele-ments around any closed circuit loop must be zero =⇒ theloop rule =⇒ this is n othing more than the conservationof energy (see Figure 18.14 in your textbook).i) Going across a battery f rom + to − is equivalent tosaying you are going in th e −∆V or −E direction.ii) Going across a battery from − to + is eq uivalentto going in the +∆V or +E direction.iii) Going in the opposite direction to current is equiv-alent to sayin g you are going in the +IR direction.iv) Going in the same direction as current is equiva-lent to saying you are going in the −IR direction.v) Current flows from +V to −V in a circuit.Donald G. Luttermoser, ETSU IV–93. Problem- Solving Strategies for Complex DC Circuits.a) To solve a particular circuit problem, you need as manyindependent equations as unknowns.b) First, draw the circuit diagram an d label all elements.Assign directions to the currents in each part of the circuit(remember, current flows from +V to −V ).c) Apply the junction rule to any junction in the circuit.d) Now, apply Kirchhoff’s loop rule to all loops in the circuit.e) Solve the simultaneous set of equations for the unkn ownquantities [realizing the point made in (a)].Example IV–2. Problem 18.27 (Page 643) from the Ser-way & Vuille textbook: (a) Can the ci rcuit shown in Figure P18.27(see diagram below ) be reduced to a single