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**Lecture - 7 Circuit Theorems **

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**Outline Source transformation. Thevenin and Norton Equivalents.**

Maximum Power Transfer. Superposition

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**Source transformation**

Source transformations allow us to exchange a voltage source (vs) and a series resistor (R) for a current source (is) and a parallel resistor (R) and vice versa. The combinations must be equivalent in terms of their terminal voltage and current. Terminal equivalence holds provided that:

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Example 1 a) For the circuit shown, find the power associated with the 6 V source. b) State whether the 6 V source is absorbing or delivering the power calculated in (a). a) We must reduce the circuit in a way that preserves the identity of the branch containing the 6 V source. We have no reason to preserve the identity of the branch containing the 40 V source. The circuit reduction is shown in the following figures:

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Example 1 The current in the direction of the voltage drop across the 6 V source is ( )/16, or A. Therefore the power associated with the 6 V source is: p6V = (0.825)(6) = 4.95 W. b) The voltage source is absorbing power.

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**Thevenin and Norton Equivalents**

Thevenin equivalents and Norton equivalents allow us to simplify a circuit comprised of sources and resistors into an equivalent circuit The equivalent circuit will be consists of a voltage source and a series resistor (Thevenin) or a current source and a parallel resistor (Norton).

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**Thevenin and Norton Equivalents**

The simplified circuit and the original circuit must be equivalent in terms of their terminal voltage and current. Thus keep in mind that : (1) the Thevenin voltage (VTh) is the open-circuit voltage across the terminals of the original circuit, (2) the Thevenin resistance (RTh) is the ratio of the Thevenin voltage to the short-circuit current across the terminals of the original circuit; (3) the Norton equivalent is obtained by performing a source transformation on a Thevenin equivalent.

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Example 2 Find the Thevenin equivalent for the circuit shown. - The current labeled ix must be zero. (Note the absence of a return path for ix to enter the left-hand portion of the circuit.) The open-circuit, or Thevenin, voltage will be the voltage across the 25Ω resistor. With ix = 0, VTh = vab = (-20i)(25) = -500i - The current i is:

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Example 2 To calculate the short-circuit current, we place a short circuit across a,b. When the terminals a,b are shorted together, the control voltage v is reduced to zero. With the short circuit shunting the 25 Ω resistor, all the current from the dependent current source appears in the short, so isc= -20i As the voltage controlling the dependent voltage source has been reduced to zero, the current controlling the dependent current source is: i=5/2000= 2.5mA

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Example 2 - Combining these two equations yields a short-circuit current of : isc = -20(2.5) = -50 mA. - From isc and VTh we get:

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**Maximum Power Transfer**

Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL. Maximum power transfer occurs when RL = Rth, the Thevenin resistance as seen from the resistor RL. The equation for the maximum power transferred is:

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Example 3 For the circuit shown, find the value of RL that results in maximum power being transferred to RL. Calculate the maximum power that can be delivered to RL c) When RL is adjusted for maximum power transfer, what percentage of the power delivered by the 360 V source reaches RL.

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Example 3 The Thevenin voltage for the circuit to the left of the terminals a,b is The Thevenin resistance is Replacing the circuit to the left of the terminals a,b with its Thevenin equivalent gives us the circuit shown, which indicates that RL must equal 25 Ω for maximum power transfer.

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**Example 3 b) The maximum power that can be delivered to RL is:**

c) When RL equals 25 Ω, the voltage vab is: - when vab equals 150 V, the current in the voltage source in the direction of the voltage rise across the source is: - Therefore, the source power is ps=-is(360) = -2520W - The percentage of the source power delivered to the load is

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Superposition The superposition principle states that whenever a linear system is excited, or driven, by more than one independent source of energy, the total response is the sum of the individual responses. In a circuit with multiple independent sources, superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are active. The current source is replaced with an open circuit and the voltage source by short circuit. Dependent sources are never deactivated when applying superposition.

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Example 4 Use the principle of superposition to find vo in the circuit. We begin by finding the component of vo resulting from the 10 V source. With the 5 A source deactivated, v'∆ must equal to (-0.4 v'∆)(10). Hence, v'∆ must be zero, the branch containing the two dependent sources is open, and v'o=10*(20/(25))= 8V

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Example 4 Then we finding the component of vo resulting from the 5 A source by deactivating the 10 V source, Then solve the resulting circuit by node voltage method. Summing the currents away from node a yields: Summing the currents away from node b gives: We now use vb = 2i’’a+ v’’∆ Therefore v’’∆= 10V and v’’o = 16V The value of vo is the sum of v’o and v’’o or 24 V.

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Summary Source transformations allow us to exchange a voltage source (vs) and a series resistor (R) for a current source (is) and a parallel resistor (R) and vice versa. Thevenin equivalents and Norton equivalents allow us to simplify a circuit comprised of sources and resistors into an equivalent circuit. Maximum power transfer is a technique for calculating the maximum value of p that can be delivered to a load, RL. superposition allows us to activate one source at a time and sum the resulting voltages and currents to determine the voltages and currents that exist when all independent sources are active.

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