Help with this please.......................

[JAWS]

Been absolutely fine with these but this one has done me.

The power dissipated in this parallel circuit, I am fine finding the power dissipated from each resistor but come a bit stuck on this.
24 ohm
100v 24 ohm
24 ohm

The way I have tried is.... 1\24+1\24+1\24=1.125 then 1\1.125=0.8

P=V2\R 1000\ 24 = 41.46x3=125 now I am lost or completely ballsed it up. Any help would be appreciated.

 

[Pete999]

Can you not provide a circuit diagram? I.m assuming you have 3 x 24ohm resistors in parallel with 100 Volts as the supply is that correct? What is it you are actually asked to find? because your post is a bit uncommunicative. As an add onit's been a very long time sibce I had to do calcs like this.

 

[JAWS]

I tried to put the circuit pic on here but its not having any of it.

Its a normal circuit with 3 resistors in parallel, each resistor is 24 ohm and the voltage is 100v. Trying to find the power dissipated in the circuit.

 

[telectrix]

Rt =8.0 ohms ( 3 x 24 in parallel )

I = V/R = 100/8= 12.5A

P = VI = 100 x 12.5 = 1250 watts.

 

[JAWS]

Rt =8.0 ohms ( 3 x 24 in parallel )

I = V/R = 100/8= 12.5A

P = VI = 100 x 12.5 = 1250 watts.

Thank you so surely their answer must be wrong.

Another question is exactly the same but with 36 ohms rather then 24ohms then the answer should be....

Rt 12.04

I=V/R = 100/12.04=8.3A

P=VI =100 x 8.3A = 830w

Yet the reason I am getting mixed up is the answer they gave is 350w

 

[hightower]

Not sure how you got Rt as 12.04. Where'd the .04 come from?

 

[Pete999]

Not sure how you got Rt as 12.04. Where'd the .04 come from?

Beat me too it HT

 

[JAWS]

Good point.

I done 1/36 + 1/36 + 1/36 = 0.083 then 1/0.083=12.04

 

[SparkyChick]

The .04 is a rounding error probably. This is why it's handy if you can work in fractions until the very last moment.

1/36 + 1/36 + 1/36 = 3/36 = 1/12

Which, when you 1/(1/12) = 12 exactly.

 

[SparkyChick]

@Massive1 ...

When dealing with parallel networks you can do away with all the 1/a + 1/b etc. malarky providing all the resistors in the network are the same value using r/n where r is the value of one of the identical resistors and n is the number of resistors in the network.

 

[telectrix]

@Massive1 ...

When dealing with parallel networks you can do away with all the 1/a + 1/b etc. malarky providing all the resistors in the network are the same value using r/n where r is the value of one of the identical resistors and n is the number of resistors in the network.

egggsackly. 2 resistors the same value, divide by 2 , 3 divide by 3 ........ 10 divide by 10.

 

[SparkyChick]

egggsackly

That made me think this... "Accrington Stanley... who are they?"