Outline Week 5: Circuits Course Notes: 3.5 Goals: Use linear algebra to determine voltage drops and branch currents.
Components in Resistor Networks voltage source current source resistor
Components in Resistor Networks 9V voltage source current source resistor
Components in Resistor Networks 9V voltage source 3A current source (inductor at an instant) resistor
Components in Resistor Networks 9V voltage source 3A current source (inductor at an instant) 6Ω resistor
V = IR +
V = IR 10V +
V = IR 10V + I = 5A
V = IR
V = IR 5A +
V = IR 5A + V = 10 (voltage drop of 10 Volts across resistor)
V = IR 0V 5A + V = 10 (voltage drop of 10 Volts across resistor)
V = IR 0V 5A + 10V V = 10 (voltage drop of 10 Volts across resistor)
V = IR Setup: Given: Resistance of resistors; voltage across voltage sources; current through current sources. Find: currents through each resistor and each voltage source; voltage drops across each current source
Kirchhoff s Laws 1. The sum of voltage drops around any closed loops in the network must be zero. 2. For any node, current in equals current out 0V 10V
40V v 1 40V 5Ω 10Ω v 2 0V
40V v 1 40V I 1 5Ω 10Ω I 2 I 3 v 2 0V
40V 40V I 1 5Ω v 1 10Ω I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 I 2 I 3 v 2 0V
40V 40V 0V I 1 v 1 5Ω I 3 I 2 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = 120 19 6.3 I 2 = 80 19 4.2 I 3 = 40 19 2.1 v 1 = 520 19 27.4 v 2 = 120 19 6.3
40V 40V 0V i 1 5Ω i 2 v 1 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = 120 19 6.3 I 2 = 80 19 4.2 I 3 = 40 19 2.1 v 1 = 520 19 27.4 v 2 = 120 19 6.3
40V 40V 0V i 1 5Ω i 2 v 1 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = 120 19 6.3 I 2 = 80 19 4.2 I 3 = 40 19 2.1 v 1 = 520 19 27.4 v 2 = 120 19 6.3 1i 1 40 + 2i 1 + 5(i 1 i 2 ) = 0 10i 2 + 5(i 2 i 1 ) = 0
40V 40V 0V i 1 5Ω i 2 v 1 v 2 I 1 = I 2 + I 3 40 v 1 = I 1 2 v 2 0 = I 1 1 v 1 v 2 = 5I 2 v 1 v 2 = 10I 3 10Ω I 1 = 120 19 6.3 I 2 = 80 19 4.2 I 3 = 40 19 2.1 v 1 = 520 19 27.4 v 2 = 120 19 6.3 i 1 = 120 19, i 2 = 40 19 1i 1 40 + 2i 1 + 5(i 1 i 2 ) = 0 10i 2 + 5(i 2 i 1 ) = 0
25Ω 30Ω 10V 50Ω 55Ω
25Ω i 2 30Ω 10V i 1 50Ω i 3 55Ω
25Ω i 2 30Ω 10V i 1 50Ω i 3 55Ω i 1 0.2449, i 2 0.1114, i 3 0.1166
Equations from previous slide: i 1 loop: 10 + i 1 + 25(i 1 i 2 ) + 50(i 1 i 3 ) = 0 i 2 loop: 25(i 2 i 1 ) + 30i 2 + (i 2 i 3 ) = 0 i 3 loop: 50(i 3 i 1 ) + (i 3 i 2 ) + 55i 3 = 0 76i 1 25i 2 50i 3 = 10 25i 1 + 56i 2 i 3 = 0 50i 1 i 2 + 106i 3 = 0
10V 3Ω 4Ω 5Ω 10V
10V 3Ω i 2 i 4 i 1 4Ω i 3 5Ω 10V
10V 3Ω i 2 i 4 i 1 4Ω i 3 5Ω 10V i 1 6.2321 i 2 3.4821 i 3 4.5357 i 4 2.6071
Equations from Previous Slide: i 1 loop: 10 + 2(i 1 i 4 ) + (i 1 i 2 ) = 0 i 2 loop: 2i 2 + (i 2 i 1 ) + 4(i 2 i 3 ) = 0 i 3 loop: 10 + 4(i 3 i 2 ) + 3(i 3 i 4 ) = 0 i 4 loop: 5i 4 + 3(i 4 i 3 ) + 2(i 4 i 1 ) = 0 3i 1 i 2 + 0i 3 2i 4 = 10 i 1 + 7i 2 4i 3 + 0i 4 = 0 0i 1 4i 2 + 7i 3 3i 4 = 10 2i 1 + 0i 2 3i 3 + 10i 4 = 0
5Ω + 40V 2A 10Ω
5A 3Ω 10V
5A i 1 3Ω i 2 10V i 3
5A i 1 3Ω i 2 10V i 3 i 1 = 5, i 2 = 0, i 3 = 25 4
10V 3Ω 5A
10V i 1 3Ω i 2 5A i 3
10V i 1 3Ω + i 2 5A i 3 Let E be the voltage drop across the current source.
10V i 1 3Ω + i 2 5A i 3 Let E be the voltage drop across the current source. i 1 = 10, i 2 = 5, i 3 = 10, E = 10
Equations from previous slide: Current Source: 5 = i 3 i 2 i 1 Loop: 10 + 3(i 1 i 3 ) + 2(i 1 i 2 ) = 0 i 2 Loop: 2(i 2 i 1 ) + E = 0 i 3 Loop: E + 3(i 3 i 1 ) + i 3 = 0 0i 1 i 2 + i 3 + 0E = 5 5i 1 2i 2 3i 3 + 0E = 10 2i 1 + 2i 2 + 0i 3 + E = 0 3i 1 + 0i 2 + 4i 3 E = 0
3Ω 4Ω 5A 10V 8A
3Ω i 1 i 2 4Ω 5A i 3 10V 8A
3Ω i 1 i 2 4Ω 5A + E 1 i 3 10V E 2 + 8A
3Ω i 1 i 2 4Ω 5A + E 1 i 3 10V E 2 + 8A i 1 8.8571, i 2 4.1429, i 3 3.8571, E 1 52.5714, E 2 42.5714
Equations from previous slide: 5A Current Source: i 3 i 1 = 5 8A Current Source: i 2 i 3 = 8 i 1 Loop: 3i 1 + 2(i i i 2 ) + E 1 = 0 i 2 Loop: 2(i 2 i 1 ) + 4i 2 E 2 = 0 i 3 Loop: E 1 + E 2 + 10 = 0 i 1 + 0i 2 + i 3 + 0E 1 + 0E 2 = 5 0i 1 + i 2 i 3 + 0E 1 + 0E 2 = 8 5i 1 2i 2 + 0i 3 + E 1 + 0E 2 = 0 2i 1 + 6i 2 + 0i 3 + 0E 1 E 2 = 0 0i 1 + 0i 2 + 0i 3 E 1 + E 2 = 10
4Ω 20V 10A 5A 4Ω
4Ω 20V i 1 + i 2 + i 3 10A 5A 4Ω
4Ω 20V i 1 + i 2 + i 3 10A 5A 4Ω i 1 = 13A, i 2 = 3A, i 3 = 2A, E 1 = 20V, E 2 = 4V Current across voltage source: 13A, top to bottom
Equations from previous slide: 10A Current Source: i 2 i 1 = 10 5A Current Source: i 3 i 2 = 5 i 1 Loop: 20 + E 1 = 0 i 2 Loop: 4i 2 + E 2 + 4i 2 E 1 = 0 i 3 Loop: 2i 3 E 2 = 0 i 1 + i 2 + 0i 3 + 0E 1 + 0E 2 = 10 0i 1 i 2 + i 3 + 0E 1 + 0E 2 = 5 0i 1 + 0i 2 + 0i 3 + E 1 + 0E 2 = 20 0i 1 + 8i 2 + 0i 3 E 1 + E 2 = 0 0i 1 + 0i 2 + 2i 3 + 0E 1 E 2 = 0
10Ω 10Ω 15A 10Ω 10Ω 10Ω
10Ω i 1 10Ω i 2 15A 10Ω i 5 10Ω i 3 10Ω i 4 clockwise: i 1 = 7.5, i 2 = 0.625, i 3 = 0, i 4 = 0.625, i 5 = 7.5
5Ω 10A 7Ω What voltage should the voltage source have, in order for there to be no current across it?
What voltage should the voltage source have, in order for there to be no current across it? R a R 1 R2 10A R3 R 4 R b
5A 5Ω 6Ω 4Ω 5A 3Ω What resistance should the top resistor have, if you want each wire touching the centre to have current 5A?
15A XXXX 10Ω 10Ω 10Ω Replace ONE resistor (with a different resistor or a different component) so that the current through the marked resistor is zero. (OK fine, one way is to remove the marked resistor itself. Try something else :) )
15A XXXX 10Ω 10Ω 10Ω Find all ways to change the resistances of the non-marked resistors so that the current flowing through the marked resistor is zero. Justify your answer with algebra.