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Linear Circuit Analysis Linear Circuits Superposition Linear Circuits • Most circuits we will study are linear • Linear circuits contain linear elements – those that have a linear relationship between their voltage and their current – Resistors – Voltage and Current Sources – Dependent sources that depend on a voltage or current (but not if they depend on a product of current and voltage or some current or voltage to a power different than one) Superposition • Linear circuits may be analyzed by looking at the voltages and currents produced by each independent source when acting alone and then adding them together • When one source acts on a circuit, the other independent sources must be set to zero – A voltage source set to zero is equivalent to a short circuit – A current source set to zero is equivalent to an open circuit Sources when set to Zero • Voltage Sources + Is Ξ Vs=0 - • Current Sources Short Circuit Ξ Open Circuit Superposition • Superposition allows us to reduce the complexity of a circuit • Superposition requires that more circuits be analyzed (more simpler circuits) • Superposition allows us to analyze circuits that contain sources of different types – DC, AC, different frequency AC sources, triangular waves, square waves, etc. Circuit Analysis using Linearity • In a linear circuit with one independent source, if you double the value of the source, then every current and voltage in the circuit doubles • Therefore, you can guess at any voltage or current in the circuit and see what the source would have to be in order to get that current and then apply a scale factor Linear Circuit Analysis • When a circuit has one independent source, you can analyze it by choosing a current or voltage somewhere in the circuit and then determine what the source had to be in order to get that value • Then you can apply a scale factor to all the assumed currents and voltages in the circuit to get the actual values of the source and other voltages and currents Example Let the 4 A current source act alone, by zeroing out the voltage source (effectively making it a short circuit. Example Short Short circuit Circuit Let the 4 A current source act alone, by zeroing out the voltage source (effectively making it a short circuit). Example Is Short Short circuit Circuit Let the current source be Is, and choose a convenient value for I4 of 10 A, then VB = 25 volts and I3 = 6.25 A Since I2=I4+I3, then I2 = 16.25 A Example Is If I2 = 16.25 A, and VA-VB = I2∙2 Ω Since VB=25 v, then VA=57.5 v Then I1=VA/14Ω= 4.11 A So Is = I1+ I2 = 20.36 A Short Short circuit Circuit Example Is Short Short circuit Circuit But Is = 4 A, so if we multiply everything by a scale factor of 4/20.36 we will get the correct voltages and currents. So I4 = 10 A ∙ (4 / 20.36) = 1.96 A Linear Circuit Analysis • When a circuit has one independent source, you can analyze it by choosing a current or voltage somewhere in the circuit and then determine what the source had to be in order to get that value • Then you can apply a scale factor to all the assumed currents and voltages in the circuit to get the actual values of the source and other voltages and currents Superposition Example Is Short Short circuit Circuit •We found if I4=10 A, Is=20.36 A, VB = 25 volts, I3 = 6.25 A, I2 = 16.25 A, VA=57.5 v & I1= 4.11 A •Knowing that Is = 4 A, we can multiply each voltage and current by 4/20.36, so Is=4 A, I4=1.96 A, VB = 4.91 v, I3 = 1.23 A, I2 = 3.19 A, VA= 11.30 v & I1= .81 A Example Open Circuit To finish the analysis by superposition, let the 34 v source act alone, by zeroing out the current source (effectively making it an open circuit. Example Open Circuit Vs •Pretend that the source is Vs and choose a convenient value for such as I1 = -I2 = 1 A. •So VA = 14 volts and VB = 16 volts and I3 = 4 A and I4 = I2 - I3 = -5 A Example Open Circuit Vs •If I4 = -5 A, then VB-Vs = I4∙2.5 Ω = -12.5 v But VB= 16 v, so Vs = 28.5 volts •Vs is really 34 v, so our scale factor is 34/28.5, so I4 = -5 A∙(34/28.5) = -5.96 A Example •If I4 = 1.96 A when the 4 A source was acting alone and if I4 = -5.96 A when the 34 volt source was acting alone, •Then I4 = 1.96 - 5.96 = -4.00 A when both sources are working together Summary • Linear circuits allow us to analyze the response to each independent source and add them together (superposition) – Especially valuable for different types of sources • Linear circuit analysis (where a current or voltage is assumed in order to get a scale factor to get the true voltages and currents) can be used for any circuit with a single independent source