Lectures_Lect09

Lectures_Lect09 - Physics 212 Lecture 9 Electric Current...

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Unformatted text preview: Physics 212 Lecture 9 Electric Current Physics 212 Lecture 9, Slide 1 Conductors Charges free to move What Determines How They Move? Spheres Cylinders Infinite Planes They move until E = 0 !!! E = 0 in conductor determines charge densities on surfaces Gauss' Law Field Lines & Equipotentials b Wa b = F dl = qE dl a a Work Done By E Field b Capacitor Networks s alls ntiia ent ote pot uiip Equ Eq F ield Li ne s Change in Potential Energy U a b = -Wa b = - qE dl a 05 b Series: (1/C23)=(1/C2)+(1/C3) Parallel C123 = C1 + C23 Physics 212 Lecture 9, Slide 2 I A Observables: V = EL I = JA t e: No R = Resistance = 1/ is , I, w w" flo flo s nt Conductivity high for good conductors. V e t r on rr c e cu J l= E l Ohm's Law: n e na i o tio ct en ire d nv I/A = V/L o I = V/(L/ ) "C L to i te os pp o I = V/R R= L A Physics 212 Lecture 9, Slide 3 Analogy to flow of water I is like flow rate of water V is like pressure R is how hard it is for water to flow in a pipe To make R big, make L long or A small L R = A To make R small, make L short or A big Physics 212 Lecture 9, Slide 4 Checkpoint 1a Same current through both resistors Compare voltages across resistors Checkpoint 1b L R A L V = IR A A2 = 4 A1 V2 = 1 V1 4 L2 = 2 L1 V2 = 2V1 Physics 212 Lecture 9, Slide 5 The SAME amount of current I passes through three different resistors. R2 has twice the crosssectional area and the same length as R1, and R3 is three times as long as R1 but has the same cross-sectional area as R1. Preflight 12 In which case is the CURRENT DENSITY through the resistor the smallest? A. Case 1 B. Case 2 C. Case 3 I J A Same Current J1 = J 3 = 2J 2 1 J A Physics 212 Lecture 9, Slide 6 Resistor Summary Series R1 R2 Parallel R1 R2 Wiring Voltage Current Each resistor on the same wire. Different for each resistor. Vtotal = V1 + V2 Same for each resistor Itotal = I1 = I2 Increases Each resistor on a different wire. Same for each resistor. Vtotal = V1 = V2 Different for each resistor Itotal = I1 + I2 Decreases 1/Req = 1/R1 + 1/R2 Physics 212 Lecture 9, Slide 7 Resistance R = R + R eq 1 2 Three resistors are connected to a battery with emf V as shown. The resistances of the resistors are all the same, i.e. R1= R2 = R3 = R. Checkpoint 2a Compare the current through R2 with the current through R3: A. I2 > I3 B. I2 = I3 C. I2 < I3 R2 in series with R3 Current through R2 and R3 is the same I 23 = V R2 + R3 Physics 212 Lecture 9, Slide 8 R1 = R2 = R3 = R Compare the current through R1 with the current through R2 Checkpoint 2b Compare the voltage across R2 with the voltage across R3 Checkpoint 2c Checkpoint 2d Compare the voltage across R1 with the voltage across R2 I1 I2 V2 V3 V1 V2 Physics 212 Lecture 9, Slide 9 I1 I23 R1 = R2 = R3 = R Compare the current through R1 with the current through R2 A. I1 /I2 = 1 /2 B. I1 /I2 = 1 /3 C. I1 /I2 = 1 D. I1 /I2 = 2 E. I1 /I2 = 3 Checkpoint 2b We kno w: I 23 = V R2 + R3 V R1 I1 R2 + R3 = =2 I 23 R1 Physics 212 Lecture 9, Slide 10 S im ila rly : I1 = V2 V23 V3 R1 = R2 = R3 = R Compare the voltage across R2 with the voltage across R3 A B C D V2 > V3 V2 = V3 = V V2 = V3 < V V2 < V3 Checkpoint 2c Consider loop V23 = V V23 = V2 + V3 R2 = R3 V2 = V3 V V2 = V3 = 2 Physics 212 Lecture 9, Slide 11 V1 V23 R1 = R2 = R3 = R Checkpoint 2d Compare the voltage across R1 with the voltage across R2 A B C D E V1 = V2 = V V1 = V2 = V V1 = 2V2 = V V1 = V2 = 1/5 V V1 = V2 = V R1 in parallel with series combination of R2 and R3 V 1= V23 R2 = R3 V2 = V3 V1 = 2V2 = V Physics 212 Lecture 9, Slide 12 V23 = V2 + V3 = 2V2 R1 V R2 Calculation In the circuit shown: V = 18V, R1 = 1, R2 = 2, R3 = 3, and R4 = 4. R3 R4 What is V2, the voltage across R2? Analysis: Ohm's Law: when current I flows through resistance R, the potential drop V is given by: V = IR. Resistances are combined in series and parallel combinations Rseries = Ra + Rb (1/Rparallel) = (1/Ra) + (1/Rb) Physics 212 Lecture 9, Slide 13 R1 V R2 Calculation In the circuit shown: V = 18V, R1 = 1, R2 = 2, R3 = 3, and R4 = 4. R3 R4 What is V2, the voltage across R2? Combine Resistances: R1 and R2 are connected: (A) in series (B) in parallel (C) neither in series nor in parallel (A) (B) (C) Parallel Combination Ra Series Combination Ra Rb Rb Parallel: Can make a loop that contains only those two resistors Series : Every loop with resistor 1 also has resistor 2. Physics 212 Lecture 9, Slide 14 R1 V R2 Calculation In the circuit shown: V = 18V, R1 = 1, R2 = 2, R3 = 3, and R4 = 4. R3 R4 What is V2, the voltage across R2? We first will combine resistances R2 + R3 + R4: Which of the following is true? (A) R2, R3 and R4 are connected in series (B) R2, R3, and R4 are connected in parallel (C) R3 and R4 are connected in series (R34) which is connected in parallel with R2 (D) R2 and R4 are connected in series (R24) which is connected in parallel with R3 (E) R2 and R4 are connected in parallel (R24) which is connected in parallel with R3 Physics 212 Lecture 9, Slide 15 R1 V R2 Calculation In the circuit shown: V = 18V, R1 = 1, R2 = 2, R3 = 3, and R4 = 4. R3 R4 What is V2, the voltage across R2? R2 and R4 are connected in series (R24) which is connected in parallel with R3 Redraw the circuit using the equivalent resistor R24 = series combination of R2 and R4. R1 V R2 4 V R1 R3 R3 R2 4 V R1 R3 R2 4 (A) (B) (C) Physics 212 Lecture 9, Slide 16 R1 V Calculation In the circuit shown: V = 18V, R3 R24 R1 = 1, R2 = 2, R3 = 3, and R4 = 4. What is V2, the voltage across R2? Combine Resistances: What is the value of R234? R2 and R4 are connected in series = R24 R3 and R24 are connected in parallel = R234 (A) R234 = 1 (B) R234 = 2 (C) R234 = 4 (D) R234 = 6 (A) (B) (C) (D) R2 and R4 in series (1/Rparallel) = (1/Ra) + (1/Rb) R24 = R2 + R4 = 2 + 4 = 6 1/R234 = (1/3) + (1/6) = (3/6) -1 R234 = 2 Physics 212 Lecture 9, Slide 17 R1 V Calculation In the circuit shown: V = 18V, R234 R1 = 1, R2 = 2, R3 = 3, and R4 = 4. I1 = I234 R24 = 6 R234 = 2 What is V2, the voltage across R2? R1 and R234 are in series. R1234 = 1 + 2 = 3 O u r n e xt ta s k is to c a lc ula te th e to ta l c urre nt in th e c irc uit Ohm's Law tells us I1234 = V/R1234 = 18 / 3 = 6 Amps V = I1234 R1234 Physics 212 Lecture 9, Slide 18 V I1234 R1234 In the circuit shown: V = 18V, R1 = 1, R2 = 2, R3 = 3, and R4 = 4. R24 = 6 R234 = 2 6A What is V2, the voltage across R2? R1 V = I1 = I234 b a I1234 = I234 = I1234 Since R1 in series w/ R234 R234 V234 = I234 R234 = 6 x 2 What is Vab, the voltage across R234 ? = 12 Volts (A) Vab = 1 V (B) Vab = 2 V (C) Vab = 9 V (D) Vab = 12 V (E) Vab = 16 V (A) (B) (C) (D) (E) Physics 212 Lecture 9, Slide 19 R 1 Calculation R1 V R234 V V = 18V R1 = 1 R2 = 2 R3 = 3 R3 R24 R4 = 4. Which of the following are true? A) V234 = V24 B) I234 = I24 C) Both A+B D) None R24 = 6 R234 = 2 I1234 = 6 Amps I234 = 6 Amps V234 = 12V Wh a t is V 2 ? R3 and R24 were combined in parallel to get R234 Voltages are same! Ohm's Law I24 = V24 / R24 = 12 / 6 = 2 Amps Physics 212 Lecture 9, Slide 20 Calculation R1 V V I1234 R1 I24 R2 V = 18V R1 = 1 R2 = 2 R3 = 3 R4 = 4. R3 R24 R3 R4 Which of the following are true? A) V24 = V2 B) I24 = I2 C) Both A+B D) None R2 and R4 where combined in series to get R24 Currents are same! Ohm's Law The Problem Can Now Be Solved! V2 = I2 R2 = 2 x 2 = 4 Volts! R24 = 6 R234 = 2 I1234 = 6 Amps I234 = 6 Amps V234 = 12V V24 = 12V I24 = 2 Amps What is V2? Physics 212 Lecture 9, Slide 21 I1 V R1 I3 R3 I2 R2 Quick FollowUps R1 a = R4 V R234 b V = 18V R1 = 1 R2 = 2 R3 = 3 R4 = 4. What is I3 ? (A) I3 = 2 A (B) I3 = 3 A (C) I3 = 4 A (A) (B) (C) V3 = V234 = 12V I3 = V3/R3 = 12V/3 = 4A R24 = 6 R234 = 2 V234= 12V V2 = 4V I24 = 2 Amps I1234 = 6 Amps What is I1 ? We know I1 = I1234 = 6 A NOTE: I2 = V2/R2 = 4/2 = 2 A I1 = I2 + I3 Make Sense??? Physics 212 Lecture 9, Slide 22 ...
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