# Physics electricity and magnetism

#### Mikee32

Joined Mar 13, 2016
9
Electricity and magnetism, please guys it's urgent is 3b right if not pls tell me how to solve it
Posted question and my work

#### WBahn

Joined Mar 31, 2012
25,899
First off, just because you put things off to the last minute does not make it urgent for anyone else on the planet.

One thing that you need to do is develop the ability to check your own work -- because in the "real world", there is no solutions manual or online forum to do it for you.

You are claiming that the needed load resistance is 7.14285 Ω. So if this is connected to a circuit having a voltage source of 7.2 V and a resistance of 6 kΩ, what is the voltage across it?

Well, let's see -- the total resistance is 6007.14285 Ω so the current is 7.2 V / 6007.14285 Ω which is 1.1986 mA, so the voltage across it is (1.1986 mA)(7.14285 Ω) which is 8.56 mV.

Seems a long way off from 5V.

Now, assuming that you meant 7.14285 kΩ (see what a difference units make?), then what do you get?

Hint, ask whether your answers make sense. If I have a voltage source with two equal valued resistors then I know that the voltage across each resistor is half the supply voltage. If you have a 6 kΩ and a 7.2 kΩ resistor, these are roughly the same size, so the voltage across the larger of the two is going to be a bit more than half of the 7.2 V supply voltage. Does it seam reasonable that this could be 5V when half the voltage is just 3.6 V?

You need to start tracking your units properly instead of just tacking on (when you DO tack on) the units that you want the answer to have.

#### Mikee32

Joined Mar 13, 2016
9
the 5V was given in the question as the v out i didn't create it, and i only want to know if my b is right please not a thank you for your time

#### WBahn

Joined Mar 31, 2012
25,899
I told you how to check if your answer to part (b) is right.

But since that isn't sufficient and since you now say that you only want to know if your answer is right, fine... No, it is not right.

#### Mikee32

Joined Mar 13, 2016
9
yes u told me how to check if it's write, but may you please guide me through to get the right answer, cs i know it's not right, been about 2 hours on it only.and my deadline is like in 10 hours so i still have time btw, if it was right i wont have posted it, let's be honest lol so please may you tell me how to come about i dont even want the answer just the working out, so i can do it for all the next questions similar cheers

#### WBahn

Joined Mar 31, 2012
25,899
You say that if it was right that you wouldn't have posted it, so that means that you knew it wasn't right when you posted it. So why did you say that you wanted to know if it was right (and in your previous post you said that that was all you wanted to know)?

You have a Thevenin equivalent circuit which, by definition, means that the load resistor can't tell the difference between the original circuit and its Thevenin equivalent.

Are you saying that you have no idea how to find a resistor that, when plugged into that Thevenin equivalent circuit, will develop 5 V across it?

Seems like a pretty straight-forward voltage divider problem, doesn't it?

• Mikee32

#### Mikee32

Joined Mar 13, 2016
9
found it the load is 13.63, cs after i get R as 7.14 which is the total R i put it in parallel formula, with 1/15 and get RL as 150/11 which is 13.63, putting it back in the formula to plus check it works fine, am i right ?

#### WBahn

Joined Mar 31, 2012
25,899
found it the load is 13.63, cs after i get R as 7.14 which is the total R i put it in parallel formula, with 1/15 and get RL as 150/11 which is 13.63, putting it back in the formula to plus check it works fine, am i right ?
I'll evaluate your work only after you at least supply the proper units for the above.

#### Mikee32

Joined Mar 13, 2016
9
yes sir, so we have 12v*r/(10kohms+r) r also in kohms, equals to 5 voltage, using normal maths we get 7.14285 kohms or 50/7 kohms it's the same thing, that's the R however
now putting it in the parallel equation we get 1/r=1/r1+1/r2
7/50 kohms = 1/15kohms + 1/Rl , Rl is the load we are trying to do
again using normal maths we get 1/rl =11/150 kohms so Rl is 150/11 kohms which is 13.63 Kohms

#### WBahn

Joined Mar 31, 2012
25,899
yes sir, so we have 12v*r/(10kohms+r) r also in kohms,
You don't need to state what the units of r are in since variables carry their units so r has units of resistance. That sufficient.

Just as it is perfectly fine to say that someone if 10 ft + 4 inches tall.

Where does this 10 kΩ come from? You are making things SO much more complicated than they need to be.

There is a reason that you were told to first find the Thevenin equivalent. Use IT to find the value of the load resistor.

By going back to the original circuit, after you found that the parallel combination of the 15 kΩ resistor and the load resistor needed to be (50/7) kΩ you then had to do a bunch more work to figure out the value of just the load resistor. And that's for a very simple circuit. What if that circuit had had a half dozen or more branches and resistors in it.

Had you done the same process with the Thevenin circuit you would have been done right then and there.

I suspect that you will not get full points for part (b) unless you use the Thevenin equivalent to get the answer.

#### Mikee32

Joined Mar 13, 2016
9
you will still get 13.63, yeah i do understand what ur trying to say, ur method is more reliable, using Vth and Rth, Thank you !!

#### WBahn

Joined Mar 31, 2012
25,899
you will still get 13.63, yeah i do understand what ur trying to say, ur method is more reliable, using Vth and Rth, Thank you !!
It's not so much that it is more reliable, or even that it is less work. It is that the point of this assignment is pretty clearly to demonstrate that you are able to both determine what the Thevenin equivalent circuit is and also that you can then use it to solve problems. If you do not actually use it to solve the problem, then you will probably not get full (or any) credit.