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Circuit Analysis Study Guide CS 2 pg
Introduction To Electrical And Computer Engineering (ENG 1450)
University of Manitoba
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Chapter 1 Charge = Coulombs (integral of current, area under the current curve) Current = Coulombs/Second (derivative of Charge, slope of charge, change in charge over time, dq/dt) Voltage = Joules/Coulomb Power: P=IV, P=(V)^2/R, P=R(I)^ Amp-hours = Current * time Watt-hours (Energy) = Power*time Watt-hours * $$$ = cost due to electricity
Chapter 2 Resistance = ρ(l/A) p = resistivity constant, l=length, A = cross sectional area Ohm's Law: V=IR, I=V/R, R=V/I Short Circuit = 0 ohms (Resistance), any current can flow through a short Open Circuit = ∞ ohms (Resistance), no current can flow through an open circuit Branches, nodes, loops: B = L + N - 1 Series: Two Elements share a single node. Same Current for all resistors in series, different Voltages (Voltage Division). Parallel: Two Elements connected to same two nodes. Same Voltage for all resistors in parallel, different Currents (Current Division). Linearity: V/I = R creates a perfectly linear relationship for certain circuits. Kirchhoff's Current Law (KCL): Current in=Current out Kirchhoff's Voltage Law (KVL): The sum of Voltages around a loop = 0
Voltage Division (Series Resistors): Two Resistors are considered in series if they have the same current pass through them. If we have a Vs and two resistors in series, here is the equations: - V1 = i(R1) - V2 = i(R2) - -V + V1 + V2 = 0 - -V + i(R1) + i(R2) = 0 The current through all resistors in series is the same, so using Ohm’s law: - V1 = [(R1)/(R1 + R2)]*V - V2 = [(R2)/(R1 + R2)]*V
• V = V1 + V2 = i(R1 + R2)
Current Division (Parallel Resistors):
Delta(Δ)-Wye(Y) Transformations:
Chapter 3 Nodal Analysis: Focuses on current flowing into and out of each node using KCL. Because V=I/R, we are actually going to find the node voltages in the end (v1, v2, v3, etc.). Steps for Nodal Analysis:
- Identify nodes in the circuit (use coloring method if necessary)
- Apply KCL at each node (except for the ground node)
- Solve the KCL equations using a matrix to find the unknown
node voltages.
Tips:
- IF there is a voltage source between two nonreference nodes (v1 and v2), that becomes a supernode. Treat the supernode as 1 node, and write a constraint equation using KCL (current coming in = current flowing out, AND voltage amount = v1-v2, using the + on the Vs as the positive node voltage and - on the Vs as the negative node voltage).
- If there is a voltage source (Vs) between a nonreference node (V1) and a reference node (Ground), do [(Vs-V1)/resistance]
- If there is a Voltage Source (Vs) right next to a nonreference node (v1) and it looks like it is the same voltage, it probably is! Do Nodal Analysis If the circuit contains:
- Many elements in parallel
- Current Sources
- Supernodes
- Circuits with fewer Nodes than meshes
- If the Node Voltage is what is being solved for Non-Planar Circuits
Mesh Analysis: Uses KVL to find unknown "Mesh Currents" (only applies in planar circuits). Steps for Mesh Analysis:
- Assign mesh currents to meshes
- Apply KVL to each of the meshes (loops)
- Solve the resulting equations to get the mesh currents
Tips:
- The mesh currents use KVL calculated CLOCKWISE around the mesh/loop.
- If the mesh current goes WITH the Voltage sorce (from --> (- +) -->), then the voltage source should be negative in the KVL equation.
- If the mesh current goes AGAINST the Voltage source (from --> (+ -) -->), then the voltage source should be positive in the KVL equation.
- IF there is a resistor bordering two meshes, make the 1st mesh current you are working on positive and the bordering mesh current negative.
- If there is a current source isolated to a mesh, that mesh current = the current source.
- If there is a current source bordering two meshes, that is classified as a SUPERMESH. For the constraint equation, use the mesh currents bordering the current source and their direction to determine which is positive and which is negative. If a mesh current is going the same direction as the current source, then it is positive in the constraint equation. If a mesh current is going against the current source, then it is negative in the constraint equation.
- After the constraint equation is written for the SUPERMESH, you then REMOVE (create an open circuit) the wire containing the current source and resistor it is attached to. Then, apply KVL on the remaining Mesh, using I1 and I2 accordingly (whatever resistors applied to I1 before the supermesh branch was removed, still applies to I1, and vice versa for I2). Do Mesh Analysis if the circuit contains:
- Many elements in series
- Voltage sources
- Supermeshes
- A circuit with fewer meshes than nodes
- If a branch/mesh current is what is being solved for
Chapter 4 Linearity: As Voltage goes up, Current goes up proportionally. The response of a circuit to a sum of sources will be the sum of the individual responses from each source separately. Linearity: - V=iR - k(iR) = k(V) - V= (i1 + i2)R = (i1)R + (i2)R = V1 + V
Superposition: If there are two or more independent sources there are Three ways to solve for the circuit parameters: Nodal Analysis, Mesh Analysis, Superposition - Superposition Principle states that the voltage/Current through an element in a linear circuit is the total sum of the voltages/currents through that element due to each independent source acting alone. Steps for Superposition: 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using techniques in Chapter 2/ 2. Repeat step 1 for each of the other independent sources 3. Find the total contribution (voltage or current) by adding all the contributions (voltages or currents) due to each independent source. Tips for Superposition: - To remove a current source, replace it with an open circuit (i=0). - To remove a voltage source, replace it with a short circuit (v=0).
Source Transformation: Replacing a Voltage source (Vs) in series with a resistor R by a current source (Is) in parallel with a resistor R, or vice versa. - Vs = Is*R - Is = Vs/R - The current source is directed towards the positive terminal of the voltage source - Source Transformation is not possible when R=0 or R=∞
Thevenin's Theorem (Thevenin Equivalents): When you have one variable element that you are trying to analyze, called the load. - Thevenin Voltage (Vth/Voc)= Open Circuit Voltage (R=infinite) - Thevenin Resistance (Rth)= Resistance looking into terminals a and b with all independent sources turned off. - Thevenin Current (Ith/In/Isc) = Vth/Rth = Ith - Use a voltage source in series with a resistance to replace complicated linear circuits It is possible for the result of this analysis to end up with a negative resistance. This implies the circuit is supplying power. This is reasonable with dependent sources Norton's Theorem (Norton Equivalents): Similar to Thevenin’s theorem, Norton’s theorem states that a linear two terminal circuit may be replaced with an equivalent circuit containing a resistor and a current source.
- Norton Current (In)= Short Circuit Current (Isc)
- Norton Resistance (Rn) = Thevenin Resistance (Rth)
- Norton Voltage (Vn) = [Vn = In * Rn]
- The Norton Current (In) is found by short circuiting the circuit's terminals and measuring the resulting current.
- Norton current is related to Thevenin voltage and resistance by source transformation: In = Vth/Rth
Norton vs. Thevenin
Maximum Power Transfer:
P = [(Vth)^2/4(Rth)]
Chapter 5 Concepts
- 5 Terminals found on all Op-amps: a. Inverting input b. Non-inverting input c. Output d. Positive and negative power supplies
- Output voltage of an op-amp can be found using the following equation: Vo = AVd = A(V+-V-) a. A is the Open Loop Gain, which is different from the closed loop gain.
- Feedback: The output of the op-amp is fed back to the inverting terminal, giving the op-amp "Negative Feedback".
- Voltage Saturation: The output voltage of the op-amp cannot exceed the input voltages. Therefore, when an output voltage should exceed the possible voltage range, the output remains at either the minimum or maximum supply voltage.
- Ideal Op-Amps: a. Infinite Open Loop Gain (A) b. Infinite Input Resistance (Ri) c. Zero output Resistance (Ro) d. Zero input current to the inverting/noninverting terminals (Io) e. Output current is NOT zero Input voltages on the inverting/noninverting terminals are equal Inverting Op-amp (ideal)
- Ideal Op-Amp Rules apply
- Vo = (-Rf/R1)*Vi
Equivalent circuit:
Non-inverting Op-Amp (Ideal)
- Ideal Op-Amp Rules apply
- Vo = Vi * [1 + (Rf/R1)]
Voltage Follower (Ideal, non-inverting op-amp) Vi = Vo
Summing Op-Amp Uses the inverting amplifier and several inputs (each with their own resistor), the summing amplifier can be used to create a simple digital to analog converter (DAC).
I1 = [(V1-Va)/R1] I2 = [(V2-Va)/R2] I3 = [(V3-Va)/R3]
Ia = I1 + I2 + I Ia = [(Va-Vo)/Rf]
Vo = - [(Rf/R1)*V1 + (Rf/R2)*V2 + (Rf/R3)*V3]
Difference Amplifier -Vo is proportional to the difference between the two inputs. -Va = Vb due to negative feedback.
Common Mode Rejection: A difference amplifier rejects any signal that is common to the two inputs, following this equation:
If this condition is true for a difference op-amp, the output is then:
Instrumentation Amplifier -Places 2 non-inverting amplifiers before the difference amplifier to increase the impedance of the difference amplifier
Unit Ramp Function Integration of the Unit Step Function
Chapter 8 Concepts Second-order Circuits (RLC circuits) o Start by getting initial conditions I(0) and dI(0)/dt (for parallel RLC circuits, V(0) and dV(0)/dt): - Capacitor: Open Circuit at long-term conditions. Voltage cannot change abruptly - V(0-) = V(0+) = Vo - Inductor: Short Circuit at long-term conditions. Current cannot change abruptly. - i(0-) = I(0+) = Io o Figure out which damping case the circuit will require, solve characteristic eqn. for roots, apply them in correct formula, solve for I or V: - Series RLC Circuits: § Overdamp ed (the roots are real and negative)
§ Critically Damped (roots are real and equal)
§ Underdam ped (roots are complex)
o Damping factor, Undampened Natural Frequency, damped natural Frequency (shown below in order)
o Solutions to characteristic equation:
o Also can be found using quadratic formula:
o Underdamped case roots:
Parallel RLC Circuits:
- All the same equations as series RLC, except: § Change I(t) --> V(t)
§ Vs = V(∞) § Change:
Damping Responses: Overdamped
Critically Damped
Underdamp ed
Chapter 9 - Sinusoidal Voltage: V(t) = Vm sin (ωt ± ɸ) o The function repeats every period, or every T seconds o T = 2π/ω o Frequency (Hertz): f = 1/T o Angular frequency: ω = 2πf - Complex numbers: o Rectangular form: z = x + jy o Polar form: Z = r<ɸ o Exponential form: Z = rejɸ - Rectangular to polar: § r = sqrt(x 2 +y 2 ) § ɸ = tan-1(y/x) - Polar to rectangular: § X = rcosɸ § Y = rsinɸ - Phasors o Converting from time domain to phasor: - V(t) = Vm cos (ωt ± ɸ) --> Vm<ɸ - V(t) = Vm sin (ωt ± ɸ) --> Vm<ɸ - 90° - Currents work the same way, just replace Vm with Im. o Phasor Relationships (ELI THE ICE MAN) - Resistors: V=IR § Voltage and Current are in phase with each other (0° difference) - Inductors: R = V/I = jωL § ELI --> Voltage Leads Current by 90°, or current lags voltage by 90° - Capacitors: R = -j/ωC § ICE --> Current leads voltage by 90° o Impedance --> phasor version of resistance o Admittance --> inverse of impedance, phasor version of conductance o In the rectangular form of Impedance (z = x + jy), the real part (x) is the resistance, and the imaginary part is the reactance (jy) - When the impedance is positive, it's inductive (jwl), when it's negative it's capacitive (-j/ωC) o In the rectangular form of admittance (z = x + jy), the real part (x) is the conductance, and the imaginary part is the susceptance (jy)
Chapter 10 - Steps to analyze AC Circuits: o Transform the circuit to the phasor or frequency domain o Solve the problem using circuit techniques o Transform back to time domain. - Perform the following as in DC: o Nodal Analysis o Mesh Analysis o Superposition o Source Transformation o Thevenin and Norton Equivalents o Op Amp Analysis
Chapter 11 - Instantaneous Power: (1/2)VmIm[cos(ɸv - ɸi)+cos(2ωt + ɸv + ɸi)] o The rate at which an element absorbs power, or the power at any instant in time o 1st cosine is constant power, 2nd cosine is sinusoidal power - Average Power: o (1/2)VmImcos(ɸv - ɸi) o Average power absorbed by an inductor and capacitor is zero watts.
o Average power is also: For periodic current:
For DC Current:
- Resistive vs. Reactive Power o Resistive: When ɸv - ɸi = 0°, the voltage and current are in phase and the circuit is purely resistive - P = (1/2)VmIm = (1/2)Im2R = (1/2)|I| 2 R o Reactive: When ɸv - ɸi = ± 90 °, the circuit absorbs no power and is purely reactive - P = (1/2)VmImCos(± 90 °) = 0
- Maximum Average Power o Pmax = (|VTh| 2 )/(8RTh)
- RMS o for a sinusoidal waveform, the RMS value is related to the amplitude as follows:
o RMS Power can be determines from either RMS current or voltage: - P = (IRMS) 2 R = (VRMS) 2 /R
Complex Power o S = (1/2)VI*
I* = complex conjugate of current o S = P + jQ
P = Real power -->! cos(&! − &")
jQ = Reactive power -->! sin(&! − &") o S = VRMSIRMS* o S = |VRMS||IRMS|<(ɸv - ɸi)
Apparent power o The product of RMS voltage and current will be called apparent power. - |S| = |VRMS||IRMS| = +,# + .#
Power Factor o P/S = cos(ɸv - ɸi) --> cos-1(P/S) = Power Factor Ratio (between 0 and 1) - P = Real Power - S = Apparent power - Real Power (P) = Apparent Power (S) * Power Factor (PF) - Power Factor = cos(ɸv - ɸi)
Adding a capacitor o To mitigate the inductive aspect of the load, a capacitor is added in parallel with the load. o With the same supplied voltage, the current draw is less by adding the capacitor. o / = %&$! "#$%
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Circuit Analysis Study Guide CS 2 pg
Course: Introduction To Electrical And Computer Engineering (ENG 1450)
University: University of Manitoba
