Circuit Analysis PDF - Kirchhoff's Laws & Resistors

11.6 and 11.8 Kirchhoff’s Laws and Circuits Series Circuits A series circuit consists of loads (resistances) connected in series....

11.6 and 11.8 Kirchhoff’s Laws and Circuits Series Circuits A series circuit consists of loads (resistances) connected in series. The current that leaves the battery has only one path to follow. ISource = I1 = I2 = I3 = ꞏ ꞏ ꞏ = IN Current is the same 11.6 and 11.8 Kirchhoff’s Laws and Resistors in The potential difference is shared over all loads. Circuits VSource = V1 + V2 + V3 + ꞏ ꞏ ꞏ + VN PHYSICS 11 - M.LLUPO Potential difference is shared 2 11.6 and 11.8 Kirchhoff’s Laws and Circuits 11.6 and 11.8 Kirchhoff’s Laws and Circuits Kirchhoff’s Voltage Law Kirchhoff’s Voltage Law In any complete path in an electric circuit, the total electric S eq 1 1 2 2 3 3 ・・・ N N potential increase (rise) at the source(s) is equal to the total electric S eq S 1 S 2 S 3 ・・・ S N potential decrease (drop) throughout the rest of the circuit. Series Circuits Factor IS. Source 1 2 3 n S eq S 1 2 3 ・・・ N Source 1 2 3 n Cancel IS Sources S eq eq 1 ・・・ 2 N 3 Using Ohm’s law The equivalent resistance of loads in series is the sum of the resistances of the individual loads. S eq 1 1 2 2 3 3 n n Req = R1 + R2 + R3 + ꞏ ꞏ ꞏ + RN Equivalent Resistance is sum of all PHYSICS 11 - M.LLUPO 3 PHYSICS 11 - M.LLUPO 4  11.6 and 11.8 Kirchhoff’s Laws and Circuits 11.6 and 11.8 Kirchhoff’s Laws and Circuits Simplify the circuit Practice – Kirchhoff’s Voltage Law Req = R1 + R2 + R3 + ꞏ ꞏ ꞏ + RN Equivalent Resistance is sum of all ISource = I1 = I2 = I3 = ꞏ ꞏ ꞏ = IN Series Circuits VSource = V1 + V2 + V3 + ꞏ ꞏ ꞏ + VN You can simplify the circuit by replacing all resistances with the Req = R1 + R2 + R3 + ꞏ ꞏ ꞏ + RN equivalent resistance Rseries or Req: Practice: Req  R1  R2  R3 1. Four loads (3.0 Ω, 5.0 Ω, 7.0 Ω, and 9.0 Ω) are connected in series to a 12 V battery. Given: V 12 V; R1 3.0 Ω; R2 5.0 Ω; R3 7.0 Ω; R4 9.0 Ω; a) Find the equivalent resistance of the circuit Required: Req ? Req  R1  R2  R3  R4 Req  3.0Ω  5.0Ω  7.0Ω  9.0Ω R eq  24Ω PHYSICS 11 - M.LLUPO 5 PHYSICS 11 - M.LLUPO 6 11.6 and 11.8 Kirchhoff’s Laws and Circuits 11.6 and 11.8 Kirchhoff’s Laws and Circuits Practice – Kirchhoff’s Voltage Law Practice – Kirchhoff’s Voltage Law ISource = I1 = I2 = I3 = ꞏ ꞏ ꞏ = IN Series Circuits ISource = I1 = I2 = I3 = ꞏ ꞏ ꞏ = IN Series Circuits VSource = V1 + V2 + V3 + ꞏ ꞏ ꞏ + VN VSource = V1 + V2 + V3 + ꞏ ꞏ ꞏ + VN Req = R1 + R2 + R3 + ꞏ ꞏ ꞏ + RN Req = R1 + R2 + R3 + ꞏ ꞏ ꞏ + RN 1. Four loads (3.0 Ω, 5.0 Ω, 7.0 Ω, and 9.0 Ω) are connected in series to a 12 V 1. Four loads (3.0 Ω, 5.0 Ω, 7.0 Ω, and 9.0 Ω) are connected in series to a 12 V battery. battery. Given: V 12 V; R1 3.0 Ω; R2 5.0 Ω; R3 7.0 Ω; R4 9.0 Ω; Given: V 12 V; R1 3.0 Ω; R2 5.0 Ω; R3 7.0 Ω; R4 9.0 Ω; b) the total current in the circuit c) the potential difference across the 7.0 Ω load Required: I ? Required: V3 ? V 12V V3  R3 I  7.0  0.50A V3  3.5V I  I  0.50A Req 24Ω PHYSICS 11 - M.LLUPO 7 PHYSICS 11 - M.LLUPO 8  11.6 and 11.8 Kirchhoff’s Laws and Circuits 11.6 and 11.8 Kirchhoff’s Laws and Circuits Parallel Circuits Parallel Circuits In any complete path in an electric circuit, the total electric In a closed circuit, the amount of current entering a junction is equal potential increase at the source(s) is equal to the total electric to the amount of current exiting a junction. potential decrease throughout the rest of the circuit. Very similar to the water in your house analogy. Potential difference across each of the individual loads in a parallel circuit must be the same as the total potential difference across the The sum of the currents in parallel paths must equal the current battery (V). through the source. Current is split through junctions: VSource = V1 = V2 = V3 = ꞏ ꞏ ꞏ = VN Potential Difference is the same ISource = I1 + I2 + I3 + ꞏ ꞏ ꞏ + IN Current is shared PHYSICS 11 - M.LLUPO 9 PHYSICS 11 - M.LLUPO 10 11.6 and 11.8 Kirchhoff’s Laws and Circuits 11.6 and 11.8 Kirchhoff’s Laws and Circuits Resistors in Parallel Circuits Simplify the circuit The sum of the currents in parallel paths must equal the current The inverse of the equivalent resistance for resistors connected in through the source. parallel is the sum of the inverses of the individual resistances. S 1 2 3 N 1/Req = 1/R1 + 1/R2 + 1/R3 + ꞏ ꞏ ꞏ + 1/Rn Inverse of Req is sum of all inverses Using Ohm’s law. VS V V V V  1  2  3 ...  N You can simplify the circuit by replacing all resistances with the Req R1 R2 R3 RN equivalent resistance Rparallel or Req. : Voltage is the same and it will cancel: VS V V V V 1 1 1 1 1 1  S  S  S ...  S     ...  Req R1 R2 R3 RN Req RT R1 R2 R3 RN The inverse of the equivalent resistance for resistors connected in parallel is the sum of the inverses of the individual resistances. PHYSICS 11 - M.LLUPO 11 PHYSICS 11 - M.LLUPO 12  11.6 and 11.8 Kirchhoff’s Laws and Circuits 11.6 and 11.8 Kirchhoff’s Laws and Circuits Practice – Resistors in Parallel Circuits Practice – Resistors in Parallel Circuits IS = I1 + I2 + I3 + ꞏ ꞏ ꞏ + In Parallel Circuits IS = I1 + I2 + I3 + ꞏ ꞏ ꞏ + In Parallel Circuits VS = V1 = V2 = V3 = ꞏ ꞏ ꞏ = Vn VS = V1 = V2 = V3 = ꞏ ꞏ ꞏ = Vn 1/Req = 1/R1 + 1/R2 + 1/R3 + ꞏ ꞏ ꞏ + 1/Rn 1/Req = 1/R1 + 1/R2 + 1/R3 + ꞏ ꞏ ꞏ + 1/Rn 2. A 60 V battery is connected to four loads of 3.00 Ω, 5.00 Ω, 12.0 Ω, and 15.0 Ω in 2. A 60 V battery is connected to four loads of 3.00 Ω, 5.00 Ω, 12.0 Ω, and 15.0 Ω in parallel. parallel. Given: V 60,0 V; R1 3.00 Ω; R2 5.00 Ω; R3 12.0 Ω; R4 15.0 Ω; Given: V 60,0 V; R1 3.00 Ω; R2 5.00 Ω; R3 12.0 Ω; R4 15.0 Ω; a) Find the equivalent resistance of the four combined loads. b) Find the total current leaving the battery. Required: Req ? Required: IS ? 1 1 1 1 1 1 1 1 1 VS 60.0V     R eq  (    ) 1 I  Req R1 R2 R3 R4 R1 R2 R3 R4 Req 1.46 I  41.1A 1 1 1 1 1 R eq  (    ) R eq  1.46Ω 3.00 5.00 12.0 15.0 PHYSICS 11 - M.LLUPO 13 PHYSICS 11 - M.LLUPO 14 11.6 and 11.8 Kirchhoff’s Laws and Circuits Practice – Resistors in Parallel Circuits IS = I1 + I2 + I3 + ꞏ ꞏ ꞏ + In Parallel Circuits VS = V1 = V2 = V3 = ꞏ ꞏ ꞏ = Vn 1/Req = 1/R1 + 1/R2 + 1/R3 + ꞏ ꞏ ꞏ + 1/Rn 2. A 60 V battery is connected to four loads of 3.00 Ω, 5.00 Ω, 12.0 Ω, and 15.0 Ω in parallel. Given: V 60,0 V; R1 3.00 Ω; R2 5.00 Ω; R3 12.0 Ω; R4 15.0 Ω; c) Find the current through the 12.0 Ω load. Required: I3 ? V3 VS 60.0V I3    R3 R3 12.0 I  5.00A 11.9 – Complex Circuits Analyses PHYSICS 11 - M.LLUPO 15  11.9 – Complex Circuits Analyses 11.9 – Complex Circuits Analyses Complex Circuits Practice – Complex Circuits Many practical circuits consist of loads in a combination of parallel For the circuit shown: and series connections. a) Find the equivalent resistance. Consider the parallel circuit as Group A When a circuit branches, the loads in each branch must be grouped and treated as a single, or equivalent, load before they can be used in a calculation with other loads. 1 1 1 1 R e q  RA  (  ) 1  (  ) 1 R2 R3 9.0 18 R e qA  6.0  R e q  Re qA  R1  6.0  4.0 R e q  10  PHYSICS 11 - M.LLUPO 17 PHYSICS 11 - M.LLUPO 18 11.9 – Complex Circuits Analyses 11.9 – Complex Circuits Analyses Practice – Complex Circuits Practice – Complex Circuits For the circuit shown: For the circuit shown: I S  3.0A I1  3.0A I 2  2.0A b) Find the current at the source. d) Summarize current values. Knowing the Req and V at the battery, we can calculate the current that comes from the battery: V 30.0V IS  S  R e q  10.0 ; VS  30.0V R e q 10.0  I S  3.0A I 3  1.0A c) Find the current I3. e) Summarize voltage values. We can find the V1 and VA (potential difference for I S  3.0A I 2  2.0A I1  3.0A 30V 30.V the parallel Group A. V1  I1  R1  3.0A  4.0  12V V 18V 0V 18V I3  3  VA  VS  V1  30.0V  12V  18V R3 18 I 3  1.0A 0V 18V There is another way to determine the I3. How? I 3  1.0A PHYSICS 11 - M.LLUPO 19 PHYSICS 11 - M.LLUPO 20  11.9 – Complex Circuits Analyses 11.9 – Complex Circuits Analyses Practice – Complex Circuits Practice – Complex Circuits PHYSICS 11 - M.LLUPO 21 PHYSICS 11 - M.LLUPO 22 11.9 – Complex Circuits Analyses 11.9 – Complex Circuits Analyses Practice – Complex Circuits Practice – Complex Circuits The potential difference at the source for the complex circuit is 60.0V and resistors The potential difference at the source for the have the following resistance, R1 = 12.0 Ω; R2 = 10.0 Ω; R3 = 15.0 Ω and R4 = 30.0 Ω. complex circuit is 60.0V and resistors have the a) Find the equivalent resistance. following resistance, R1 = 12.0 Ω; R2 = 10.0 Ω; Consider the parallel circuit as Group A with R2 and R3 in series within the parallel R3 = 15.0 Ω and R4 = 30.0 Ω. branch. c) Find the potential difference and the current 1 1 1 1 on each load. R eqA  (  ) 1  (  ) 1 R e qA  13.6 Ω R2  R3 R4 10  15 30 ReqA  R1  ReqA  12Ω  13.64Ω Req  25.64Ω Source Load 1 Load 2 Load 3 Load 4 b) Find current at the source. V 60.0V 28.1 V 12.8 V 19.2 V 31.9 V V 60.0V IS  S  I 2.34 A 2.34 A 1.28 A 1.28 A 1.06 A Req 25.64Ω I S  2.34A R 25.6 Ω 12.0 Ω 10.0 Ω 15.0 Ω 30.0 Ω PHYSICS 11 - M.LLUPO 23 PHYSICS 11 - M.LLUPO 24  11.9 – Complex Circuits Analyses Practice – Complex Circuits Calculate the missing currents, resistances, and the potential differences for each of the loads in the circuit. Find the equivalent resistance for the circuit and the power output for the circuit. Source Load 1 Load 2 Load 3 Load 4 Load 5 Load 6 Load 7 R 48 300 Ω 100 Ω 30 90 18 30 Ω 50 I 10.0 A 1.00 A 3 4 0.667 A 3.333 6.00 A 6 V 480 300 300 120 V 60 60 180 300 V PHYSICS 11 - M.LLUPO 25