Trainee questions to answer

[jimmyray]

Hi Pennychew, Struggling with this one a bit for some reason. I get the answer as 220 Ohms, but am a bit doubtful. Had a couple of beers, which probably doesn't help!! Cheers. Jim

 

[pennychew]

Hi Pennychew, Struggling with this one a bit for some reason. I get the answer as 220 Ohms, but am a bit doubtful. Had a couple of beers, which probably doesn't help!! Cheers. Jim

Hi, its not 220ohms. If you look back to the start of the thread its very similar to the first question posted which i struggled with for about 2 days!!.

If you label the first two resistors as R1 and R2 and the resistor by itself as R3, then use the advice given to me by Marvo "Kerchoff's law would dictate that the current through R3 would equal the sum of the currents through R1 and R2."

Stu

 

[Marvo]

A couple of helpful things to remember with this one;

Firstly the volt drop across the 20ohm resistor and the unknown resistor will be the same because they're in parallel.

Secondly the supply voltage minus the volt drop across the parallel resistors will equal the volt drop across the 5ohm resistor.

Finally the current through the 5ohm resistor will equal the sum of the currents through the two resistors in parallel (Kirchoff's Law)

 

[jimmyray]

55 ohms?

 

[jimmyray]

Scrap that 55 ohms! Had a look at the books, and I think the answer is 2.86 Ohms. Hope this is right!

 

[pennychew]

Spot on! i must have done the same thing as you wrong the first time because i got 55ohms at my first attempt as well.

Stu

 

[jimmyray]

Cheers mate, definitely a case of knowing and applying all the laws and rules. Thanks to marvo and yourself for pointing out key factors. Usually quite good with ohms law, but obviously more practice required!

 

[telectrix]

Scrap that 55 ohms! Had a look at the books, and I think the answer is 2.86 Ohms. Hope this is right!

that's what I got, so either you're right, or we're both wrong.

 

[jimmyray]

Nice one telectrix. Do you want to post the next question?

 

[Marvo]

It would be good if you can post your calculations as well as the actual answer. It might help others who are following the thread or read it some time in future.

 

[jimmyray]

No problems Marvo, here goes: Calculated the voltage across the resistor in parallel using V=IR = 0.5 x 20= 10 Volts. Subtracted that result from the supply voltage, 30V-10V=20V. Use that voltage and the 5 Ohm resistor to find out the total current: I=V/R = 20/5 = 4A. As the parallel voltages are equal (10V), can use the remaining current (4-0.5 = 3.5A), to work out the unknown resistor value: R=V/I = 10/3.5 = 2.86 Ohms (rounded up). Cheers. Jim

 

[telectrix]

No problems Marvo, here goes: Calculated the voltage across the resistor in parallel using V=IR = 0.5 x 20= 10 Volts. Subtracted that result from the supply voltage, 30V-10V=20V. Use that voltage and the 5 Ohm resistor to find out the total current: I=V/R = 20/5 = 4A. As the parallel voltages are equal (10V), can use the remaining current (4-0.5 = 3.5A), to work out the unknown resistor value: R=V/I = 10/3.5 = 2.86 Ohms (rounded up). Cheers. Jim

spot on there. jimmy. i'll think of one to post in the morning. need to be sober to do a pic.

 

[matt1386]

Ok guys ill post a question, this is to do with cells connected in series :

if six 2.2v cells were connected in series, each with a internal resistance of 0.01 ohms, what would be the terminal voltage if a current of 3A were flowing ?

Matt

 

[Chrisuk]

A question on resistivity as thats we have just been doing...

In the following exercise take the resistivity of copper as 1.78 x10-8 Ωm.

Find the cross sectional area of a copper conductor 42m long which carries a current of 36A with a voltage drop of 2.69v.

Can you explain how to work this one out because the help screen shot u put up only shows how's to work out festivity which I know how to do
If u had given the resistivity maybe I could of found the area but so far we have been told how to work out resistivity PlxL/A so I'd appreciate the working out on that one

 

[Chrisuk]

Blank!