Thevenin's Equivalent Problem 5.36

How to solve a circuit problem using a thevenin equivalent circuit.

[@chrisott8142]

I think that @ about 5:00 you should explain that a parallel branch will have the same voltage, rather than saying that the branch doesn't matter. It's a bit more confusing when someone says that something doesn't mater, rather than explaining why.

[@MJtheP]

Shit like this is exactly why I have to watch youtube videos to learn the material. My EE professor's favorite phrase is "it doesn't matter" everything in the course "doesn't matter" till he throws that damn test in front of us and suddenly it does

[@MrWoopydalan]

why do you replace the 40 V source as an open circuit? I thought voltage sources are to be replaced with short circuits

[@georgemonnatjr.172]

+MrWoopydalan Because the 40V source is the load and not a part of the Thevenin equivalent circuit.

[@baddyvinny7970]

I thought the voltage source had to be short circuited

[@arunkaurav8894]

Sir I do not understood

[@seanrogers5782]

This was extremely helpful. 9:10 - 9:13 You said resistors instead of voltage sources though. Thanks

[@AraujoMatt]

In the video, I explain that the voltage V'oc is the voltage across the 5 and 10 ohm resistors that are horizontally below the current source. For the Rth, you look at the resistance that the load sees. Because we have only an independant source, we remove it. Current sources are treated as open circuits. The drawing shows that the 5 ohm resistor is "dangling" and doesn't contribute the the resistance seen by the load.

[@Simmons101]

around 4:15 I noticed when using super position you created an open circuit where the 40 V source was, but I believe when deactivating sources, you create an open circuit for current sources and a short circuit for a voltage source. Right here it looks like you still have a voltage where the 40 V source was, but it wasn't necessarily deactivated. Wouldn't you need to deactivate one at a time and then add up their sums?

[@NickStilson]

When you were calculating V'oc, why didn't include the 5ohm resistor? For that matter why didn't you include the same resistor when you calculated Rth, why doesn't it contribute to the circuit?

[@hamodyeee]

When you calculate Req you delete all source and stop , its must convert some short circuit and open circuit not deleting and keep circuit open circuit like what you doing .

[@AH17293]

I think you consider the load to be the element you require the current or power for.

[@zawette]

if we used nodal analysis instead would we still need to use superposition

[@mogee6417]

"We're cooking with gas?"

[@SmoothChino]

cooking with electricity

[@maximumcarnage01]

How do you choose which element is considered the load? Is it arbitrary?

[@SmoothChino]

How did you get 40 +20 volts? aren't all Resistors in series on top? shouldn't it be 20 ohms * 4v?

[@tadm123]

thanks man i appreciate it.

[@KeijiJohnLibadisos]

why you remove 2A?

[@idzuanbahar9712]

very helpful. for me, your "LOAD" is "SOURCE" to me. and 1 more sir, u did not short circuited your voltage source in calculating Rth. quite confusing me. anyway. good video though. really.

[@carstuff4260]

Mistake at 5:11 ?

[@cranchianflame1007]

J Lang what mistake?

[@geloace28]

i thought that is 2A not mA

[@jikaikas]

Wtf is happenin

[@maryam.654]

We can get Voc by easy way -Voc+4x10+5x6+2x20=0- Voc=110

[@flapperkhut9586]

It must be 60V + (-50V)=10V I think...

[@baitywahidiah8077]

I got the same answer as u ..

[@MarihuanaECTOSPASM]

arent they the same polarity, so you can add them up?