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Thevenin’s Theorem-Basic Electrical Engineering-Lecture Slides, Slides of Electrical Engineering

Bengal Engineering & Science UniversityElectrical Engineering

Dr. Priya Gupta delivered this lecture at Bengal Engineering

Typology: Slides

2011/2012

Uploaded on 07/03/2012

burman
burman 🇮🇳

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Thevenin’s Theorem
–Thevenin’s theorem simplifies the process of
solving for the unknown values of voltage and
current in a network by reducing the network to
an equivalent series circuit connected to any pair
of network terminals.
–Any network with two open terminals can be
replaced by a single voltage source (VTH) and a
series resistance (RTH) connected to the open
terminals. A component can be removed to
produce the open terminals.
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Download Thevenin’s Theorem-Basic Electrical Engineering-Lecture Slides and more Slides Electrical Engineering in PDF only on Docsity!

Thevenin’s Theorem

  • Thevenin’s theorem simplifies the process of

solving for the unknown values of voltage and

current in a network by reducing the network to

an equivalent series circuit connected to any pair

of network terminals.

  • Any network with two open terminals can be

replaced by a single voltage source (VTH) and a

series resistance (RTH) connected to the open

terminals. A component can be removed to

produce the open terminals.

Thevenin’s Theorem

Fig. Application of Thevenin’s theorem. ( a ) Actual circuit with terminals A and B across RL. ( b ) Disconnect RL to find that VAB is 24V. ( c ) Short-circuit V to find that RAB is 2Ω.

Thevenin’s Theorem

• Determining Thevenin Resistance and Voltage

  • RTH is determined by shorting the voltage source and calculating the circuit’s total resistance as seen from open terminals A and B.
  • VTH is determined by calculating the voltage between open terminals A and B.

Thevenin’s Theorem

( a ) VAB is still 24V. ( b ) Now the RAB is 2 + 4 = 6 Ω. ( c ) Thevenin equivalent circuit.

Thevenizing a Circuit

with Two Voltage

Sources

Steps:

  • Disconnect R3 from the circuit
  • Since E1 and E2 oppose each other , the net voltage is E1 – E2 = 30V – 26V = 4V
  • Total Resistance is R1 + R2 = 12 + 4 = 16 ohm
  • Thus, current is 4V /16 ohm = 0.25 A (CCW)
  • ER1 = 0.25 x 12 = 3V and ER2 = 0.25 x 4 =1V

Steps:

  • Now voltage between a&b is 30 – 3 = 27 V (left)
  • Also the same 1 + 26 = 27V (right)
  • Rth = R1 // R2 = 3 ohm
  • Reconnect R3 and find current

E/Rt = 27V / 9 ohm = 3A

Steps:

  • Now voltage between a&b is 20 – 4 = 24 V (left)
  • Also the same 28-4 = 24 (right)
  • Rth = R1 // R2 = 4 ohm
  • Reconnect R3 and find current

E/Rt = 24V / 12 ohm = 2A

Thevenizing a Circuit with Two Voltage Sources

Fig. Thevenizing a circuit with two voltage sources V 1 and V 2. ( a ) Original circuit with terminals A and B across the middle resistor R 3. ( b ) Disconnect R 3 to find that VAB is −33.6V. ( c ) Short-circuit V 1 and V 2 to find that RAB is 2.4 Ω. ( d ) Thevenin equivalent with RL reconnected to terminals A and B.

Example: