Phys 102 – Lecture 8 Circuit analysis and Kirchhoff’s rules 1

Recall from last time... We solved circuits like... + R3R3 R1R1 R2R2 ε What about a circuit like... by combining series & parallel components Simplify + R tot ε Expand Phys. 102, Lecture 8, Slide 2 R1R1 + R3R3 ε2ε2 ε1ε1 +

Kirchhoff’s loop rule Voltages around a loop sum to zero Go around loop and write +V element if electric potential increases –V element if it decreases Phys. 102, Lecture 8, Slide 3 Resistors: higher/lower potential depends on current direction R Capacitors: higher/lower potential depends on which plate has +Q/–Q C + Batteries: + end is always at higher potential ε Label +/– for higher/lower electric potential Element 2 Element 4 Element 3Element 1 + – I Is voltage positive or negative? + – +Q+Q –Q

Calculation: single loop practice + R 1 = 5 Ω ε 1 = 50 V R 2 = 15 Ω ε 2 = 10 V + I +– + – What if we go around the loop the “wrong” way? What if we’re not given the current direction? What if we pick the “wrong” direction? –+ Calculate the current I in the circuit + – + – Phys. 102, Lecture 8, Slide 4

Calculation: single loop practice Phys. 102, Lecture 8, Slide 5 How can the current be driven opposite battery 2? + R 1 = 5 Ω ε 1 = 50 V R 2 = 15 Ω + I IR 1 ε1ε1 ε2ε2 IR 2 ε 2 = 10 V I I

ACT: Checkpoint 1.1 Phys. 102, Lecture 8, Slide 6 Calculate the current through R 1. + R 1 = 10 Ω ε 1 = 10 V I1I1 ε 2 = 5 V + R 2 = 10 Ω I2I2 IBIB A. I 1 = 0.5 A B. I 1 = 1.0 A C. I 1 = 1.5 A

ACT: Checkpoint 1.2 Phys. 102, Lecture 8, Slide 7 Calculate the current through R 2. + R 1 = 10 Ω ε 1 = 10 V I1I1 ε 2 = 5 V + R 2 = 10 Ω I2I2 IBIB A. I 2 = 0.5 A B. I 2 = 1.0 A C. I 2 = 1.5 A

C Nerve cell equivalent circuit Phys. 102, Lecture 8, Slide 8 V out V in Neurons have different types of ion channels (K +, Na +, and Cl – ) that pump current into and out of cell – act like batteries! R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na S V out V in

ACT: loop Phys. 102, Lecture 8, Slide 9 Na + channels have a “gate” (represented by the switch S) that allows or blocks ion flow. In its resting state, a Na + channel is shut (i.e. switch S is open). Which equation is correct? R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na S A. B. C.

Calculation: electric potential Find the electric potential difference across the cell V in – V out (Assume V out = 0 for reference) or: ABCDA V Phys. 102, Lecture 8, Slide 10 + R Cl εKεK I + RKRK ε Cl V in V out = 0 A B C D V in V out ε K = 80 mV, ε Na = 60 mV, ε Cl = 50 mV R K = 2 MΩ, R Na = 0.2 MΩ, R Cl = 5 MΩ

Kirchhoff’s junction rule Phys. 102, Lecture 8, Slide 11 The sum of currents into a junction equals the sum of currents out of a junction I2I2 I4I4 I5I5 I1I1 I3I3 Example:

ACT: Checkpoint 1.3 Phys. 102, Lecture 8, Slide 12 Calculate the current through the battery I B. + R 1 = 10 Ω ε 1 = 10 V I1I1 ε 2 = 5 V + R 2 = 10 Ω I2I2 IBIB A. I B = 0.5 A B. I B = 1.0 A C. I B = 1.5 A

Calculation: Kirchhoff’s laws In the circuit, the current through R C is 0. What is the current through R 3 and the value of R x ? + RxRx R 1 = 10 Ω RCRC ε = 12 V R 2 = 20 ΩR 3 = 30 Ω From EX1 FA13 No current flows through R C so I 2 = I 3 No current flows through R C so V C = 0 and I 1 = I x I1I1 I2I2 IxIx I3I3 Phys. 102, Lecture 8, Slide 13

Nerve cell equivalent circuit Phys. 102, Lecture 8, Slide 14 R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na During nerve impulse, Na + channels open (i.e. switch S closes) and allow Na + to enter the cell S V out = 0 V in = – 70mV ? What happens to the currents through the channels and the potential in the cell?

Calculation: two loop circuit R Na + R Cl εKεK + RKRK 1.Label all currents 2.Label +/– for all elements 3.Choose loop and direction 4.Write down voltage differences ε Cl Given the circuit to the right, find I K, I Na and I Cl and V in – V out. Phys. 102, Lecture 8, Slide 15 + ε Na ε K = 80 mV, ε Na = 60 mV, ε Cl = 50 mV R K = 2 MΩ, R Na = 0.2 MΩ, R Cl = 5 MΩ V in V out

ACT: Kirchhoff loop rule What is the correct expression for “Loop 3” in the circuit below? A. B. C. Phys. 102, Lecture 8, Slide 16 R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na Loop 3

Calculation: two loop circuit Phys. 102, Lecture 8, Slide 17 We have 3 unknowns, need 3 equations Loop 1: Loop 2: Loop 3: 5.Write down junction rule R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na Loop 3 Loop 1 Loop 2 Given the circuit to the right, find I K, I Na and I Cl and V in – V out. ε K = 80 mV, ε Na = 60 mV, ε Cl = 50 mV R K = 2 MΩ, R Na = 0.2 MΩ, R Cl = 5 MΩ V in V out

ACT: Kirchhoff junction rule What is the correct expression for junction in the circuit? A. B. C. Phys. 102, Lecture 8, Slide 18 R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na

Calculation: two loop circuit Phys. 102, Lecture 8, Slide 19 3 equations, 3 unknowns, the rest is algebra! (3) Substitute Eq. (3) into Eq. (2) and rearrange (2’) Given the circuit to the right, find I K, I Na and I Cl and V in – V out. ε K = 80 mV, ε Na = 60 mV, ε Cl = 50 mV R K = 2 MΩ, R Na = 0.2 MΩ, R Cl = 5 MΩ R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na Loop 1 Loop 2 (1) (2) V in V out

Calculation: two loop circuits Phys. 102, Lecture 8, Slide 20 Substitute I K in Eq. (1) into Eq. (2’) and rearrange Plug solution into Eq. (2’) to get I K Now 2 equations (1 and 2’), 2 unknowns (I K and I Na ) Use junction Eq. (3) to get I Cl (2’) (1)

ACT: Kirchhoff junction rule We found that I K = 62 nA, I Na = 81 nA and I Cl = –19 nA. Which of the following statements is FALSE? A. I K is out of the cell B. I Na is into the cell C. I Cl is into the cell Phys. 102, Lecture 8, Slide 21 R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na Outside Inside

Calculation: two loop circuit Phys. 102, Lecture 8, Slide 22 R Na + R Cl εKεK I Na I Cl IKIK + RKRK ε Cl + ε Na S V out = 0 V in = – 70mV Find the new V in – V out : V out = 0 V in

Summary of today’s lecture Two basic principles: Kirchhoff loop rule Voltages around circuit loop sum to zero (based on conservation of energy) Kirchhoff junction rule Currents into a circuit branch equal currents out (based on conservation of charge) Phys. 102, Lecture 8, Slide 23

Summary of today’s lecture Phys. 102, Lecture 8, Slide 24 Basic approach to solving complex circuits: 1. Label all currents 2. Label +/– for all elements 3. Choose loop(s) and direction(s) 4. Write down voltage differences 5. Write down junction rule The rest is algebra!