Blog #34 - Calculating Current using Thevenin’s Theorem

As part of my problem series in this Blog, I will be Calculating Current Using Thevenin’s Theorem.

Once I am finished with this series of problems I will be posting them in my Stan store,at this web address.

Before proceeding I want to explain the three WEB addresses that you will be directed to using. You have already seen the first…https://stan.store/GVB…this is the web address of my Stan store which will give you direct access to all of my electrical courses. 

On the last page  you'll find an address,,,HTTP://bitl.ly/47YB3vh… that will direct you to obtaining the 50 page crib sheets and notes that will not only be handy when you're taking any of the courses in my Stan store, but also for reference during any time during your career. Here, you will also be asked for your email address which will not be shared or distributed anyway but it will allow me to keep in touch and let you know of any additions or updates to my courses and blogs.

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Task Problem: Calculate the value of the current IL through the resistor RL in this dc network using Thevenin’s Theorem.

Thevenin’s theorem states that: “Any two-terminal linear network containing resistances and sources of emf and current may be replaced by a single source of emf in series with a single resistance. The emf of the single source will be called ETh, at the network terminals A & B. The single-series resistance, called RTh, is the resistance between the network terminals A & B when all of the independent sources are replaced by their internal resistances”.

When the Thevenin equivalent circuit is determined for a network, the process is known as “thevenizing” the circuit.

“Thevenizing” the circuit to find the Thevenin Voltage ETh.

The load resistor is removed. The open-circuit terminal voltage of the network is calculated;  this value is ETh. Because no current can flow through R3, the voltage ETh is the same as the voltage across resistor R2. Using the voltage-divider rule we find that…

ETh = (100 V) X [100/(100 + 100)] = 50 V.

Next when “thevenizing” the circuit we find the Thevenin Resistance RTh.

The network is redrawn with the source of emf replaced by a short circuit. (If a current source is present, it is replaced by an open circuit.) 

The resistance of the redrawn network as seen by looking back into the network from the load terminals is calculated. We call this value is RTh,

where RTh = (50 Ω) + (100 Ω)||(100 Ω) = 100 Ω.

The Thevenin equivalent circuit consists of the series combination of ETh and RTh. 

The load resistor RL is connected (or re-connected) across the output terminals of this equivalent circuit… 

RT = RTA + RTh  = 100 + 50 = 150 Ω, and 

IL =  ETH/RT = 50/150 = 1/3 A.

With respect to the terminals only, the Thevenin circuit is equivalent to the original linear network.  

Changes in RL does not require any calculations for a new Thevenin circuit. The simple series Thevenin circuit can be used to solve for load currents each time RL is changed.

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Remember, this video has been brought to you by PSPT, where you will find electrical train training videos when you go to this web address,..https://bit.ly/47YB3vh… which will also give you a free copy of my 50 page crib sheets that you can use while viewing any of the courses or just keep handy during your every day work.