Maker Pro
Maker Pro

Simple maths resistance exercice

LvW

Apr 12, 2014
604
Joined
Apr 12, 2014
Messages
604
If you have any specific question - don`t hesitate.
Up to now - I see no question.
 

Ratch

Mar 10, 2013
1,099
Joined
Mar 10, 2013
Messages
1,099
https://ibb.co/irx3yz

Only the example exercise at the end of the picture.I only know ohms law and that P=VxI
How am i supposed to do it

Find the voltage across the resistor or the current present in the resistor and compute the power. Use mesh or node analysis to find the current or voltage.

Why could you not post only the schematic in a upright display instead of the irrelevant whole page on its side. Do you read a book sideways like that?

Ratch
 

hevans1944

Hop - AC8NS
Jun 21, 2012
4,878
Joined
Jun 21, 2012
Messages
4,878
Well just do the exercise for me and post it here cause i cant do it
Sorry, @endlessparadox, we don't roll that way here. You do the exercise, you post the results here, then maybe someone will comment on your work. Do what @Ratch said in his post #4.

This is NOT a teaching forum. If you do not know how to apply Ohm's Law and Kirchoff's voltage and current laws to node and mesh analysis, learn from a text book or learn from on-line study. Do the work, and show us what work you have done, if you expect to receive any guidance. We will not do your homework for you.
 

endlessparadox

Feb 14, 2018
40
Joined
Feb 14, 2018
Messages
40
Find the voltage across the resistor or the current present in the resistor and compute the power. Use mesh or node analysis to find the current or voltage.

Why could you not post only the schematic in a upright display instead of the irrelevant whole page on its side. Do you read a book sideways like that?

Ratch
Sorry, @endlessparadox, we don't roll that way here. You do the exercise, you post the results here, then maybe someone will comment on your work. Do what @Ratch said in his post #4.

This is NOT a teaching forum. If you do not know how to apply Ohm's Law and Kirchoff's voltage and current laws to node and mesh analysis, learn from a text book or learn from on-line study. Do the work, and show us what work you have done, if you expect to receive any guidance. We will not do your homework for you.

The thing is RT=R1+R2+R3 IT=I1+I2+I3
and then ohms law.i dont have any current in the exercise.So how am i supposed to do it.if i calculate all the series ohms it makes 37 ohm if i then calculate the pararel resistors ohms it makes 5.22ohm a total of 42.22ohm.
I take the total voltage divided by the resistance for the current.it makes a total of 0.189 Amperes.Then the voltage for the 6ohm resistor which makes 6*0.189=1.134v.
Then for the power P=I*V
1.134*0.189=0.21 Watts?

Is it the answer?
Or is there some law i dont know?



******
and sorry for the image i take pictures on phone
******
 

hevans1944

Hop - AC8NS
Jun 21, 2012
4,878
Joined
Jun 21, 2012
Messages
4,878
Is it the answer?
Or is there some law i dont know?
Your "answer" isn't even close to being correct. Do you know how two (or more) resistors in series can be replaced with a single equivalent resistance? Do you know how two (or more) resistors in parallel can be replaced with a single equivalent resistance? You must reduce your circuit to a single equivalent resistance connected to the 8V battery by combining series and parallel equivalent resistances. Start with the series-connected 15Ω and 7Ω resistors, which are in parallel with the 22Ω resistor. Work from right to left until you have a single resistance connected to the 8V battery.

For two resistances, R1 and R2, in series the equivalent resistance Rseries = R1 + R2. For two resistances, R3 and R4, in parallel the equivalent resistance Rparallel = (R3)(R4) / (R3+R4). Another way to write this is 1/Rparallel = 1/R3 + 1/R4. Or in words, the reciprocal of two resistances in parallel is the sum of their reciprocal resistances.

So, if you draw your sideways circuit out so you can see it properly, and then proceed to reduce the circuit to a single equivalent resistance connected across the 8V battery, you can calculate the current this equivalent resistance draws from the 8V battery. This will be the same current that is drawn through the 6Ω resistor. Why? Because all the current drawn from the battery passes through the 6Ω resistor. The square of this current multiplied by 6Ω will tell you how much power is dissipated in the 6Ω resistor, which was the original question that was asked.
 

endlessparadox

Feb 14, 2018
40
Joined
Feb 14, 2018
Messages
40
Your "answer" isn't even close to being correct. Do you know how two (or more) resistors in series can be replaced with a single equivalent resistance? Do you know how two (or more) resistors in parallel can be replaced with a single equivalent resistance? You must reduce your circuit to a single equivalent resistance connected to the 8V battery by combining series and parallel equivalent resistances. Start with the series-connected 15Ω and 7Ω resistors, which are in parallel with the 22Ω resistor. Work from right to left until you have a single resistance connected to the 8V battery.

For two resistances, R1 and R2, in series the equivalent resistance Rseries = R1 + R2. For two resistances, R3 and R4, in parallel the equivalent resistance Rparallel = (R3)(R4) / (R3+R4). Another way to write this is 1/Rparallel = 1/R3 + 1/R4. Or in words, the reciprocal of two resistances in parallel is the sum of their reciprocal resistances.

So, if you draw your sideways circuit out so you can see it properly, and then proceed to reduce the circuit to a single equivalent resistance connected across the 8V battery, you can calculate the current this equivalent resistance draws from the 8V battery. This will be the same current that is drawn through the 6Ω resistor. Why? Because all the current drawn from the battery passes through the 6Ω resistor. The square of this current multiplied by 6Ω will tell you how much power is dissipated in the 6Ω resistor, which was the original question that was asked.
This time i got 16 volt could ypu check what i did wrong there are 3 pictures but i write big

Watch from the last photo to the first photo
https://ibb.co/hZqCAe
https://ibb.co/jLETiz
https://ibb.co/kPkeqe
 

hevans1944

Hop - AC8NS
Jun 21, 2012
4,878
Joined
Jun 21, 2012
Messages
4,878
Close but no cigar. You messed up in the last step, converting two 6Ω resistors which are in series into a 3Ω equivalent resistance. Correct this and then calculate the current again. Power in the original 6Ω resistor is the corrected current squared multiplied by six ohms.
 
Top