How to Calculate Resistance and Current in a Parallel Circuit

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To calculate resistance and current in a parallel circuit, the equivalent resistance of resistors in parallel must be determined correctly, which is not simply the sum of their resistances. The potential difference across PQ is not the same as the total voltage of 14V, as part of it is used across the parallel resistors QR. The total current through the network can be found using Ohm's law once the equivalent resistance is established. Confusion arises when trying to apply series and parallel rules without understanding their distinct configurations. Understanding these principles is crucial for accurately solving circuit problems.
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Im stuck on this question.

There is a image of a network. P-Q-R

P goes to Q, and then splits into 2 routes into R.
On each route there is a resistor.
The resistor P-Q is 5 ohms, and the resistors on the 2 different routes Q-R are 3 ohms, and the other is 6 ohms.
The p.d. across PR is 14 volts.
Calculate

a) the equivalent resistance of network QR.
Wouldnt this just be 9ohms. 6+3?

b) the current flowing through PQ.
i did 14/5 = 2.8A

c) the p.d across PQ.
really confused, why is it not just 14V

d) the current through the 3 ohm resistor.
i did 7/3 = 2.3A

I used the equation R = V/I
 
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coolnufc said:
a) the equivalent resistance of network QR.
Wouldnt this just be 9ohms. 6+3?

No. You have to consider how the resistors between Q and R are connected.

b) the current flowing through PQ.
i did 14/5 = 2.8A

No. 14V is the potential difference across PR, not PQ.

c) the p.d across PQ.
really confused, why is it not just 14V

Because that would mean that the potential difference across QR is zero, and hence no current flows through the entire network.

d) the current through the 3 ohm resistor.
i did 7/3 = 2.3A

Where does 7/3 come from?
 
Tom Mattson said:
Where does 7/3 come from?

No idea. I am really confused.
 
OK, start with a. How are those two resistors connected?
 
Its hard to explain. P is connected to Q with a resistor in the middle. then Q goes in two different routes to R. One resistor on each route. One resistor is 3ohms, then other is 6ohms. and the resistor between P and Q is 5 ohms.
 

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coolnufc said:
Its hard to explain.

No it's not, you've got 2 choices:

1.) They are connected in series.

2.) They are connected in parallel.

Can you tell me which?
 
the resistor connecting P and Q is in series.
The two resistors connecting Q and R are in parallel.
 
coolnufc said:
The two resistors connecting Q and R are in parallel.

Right. And how do you compute the equivalent resistance of two resistors in parallel?
 
Tom Mattson said:
Right. And how do you compute the equivalent resistance of two resistors in parallel?

Is the answer 2?
 
  • #10
Yes. Do you understand why?
 
  • #11
yeah i do. but I am still confused with the other questions.
why is 14V not across PQ as well as the whole thing, it hasnt split into parallel yet?
 
  • #12
coolnufc said:
why is 14V not across PQ as well as the whole thing,

Because that would mean that the potential difference across QR is zero, which means that no current flows through them. But that is impossible because of conservation of charge at junction Q. Part of the 14V is across PQ and part of it is across QR.

So for part b tell me this: What's the equivalent resistance of the entire network?
 
  • #13
Tom Mattson said:
So for part b tell me this: What's the equivalent resistance of the entire network?

Umm, well i came out with two answers. :S
I got 1.43. and then i tried another way and got 6.43.
 
  • #14
I have no idea of what you're thinking unless you show how you got those numbers (both are wrong).
 
  • #15
well on part a. i did 3 x 6 / 3 + 6. so i tried that for part b. by doing 3 x 6 x 5 / 3 + 6 + 5 = 6.43.

and then i did 1 / (1/6 + 1/5 + 1/3) = 1.43.
 
  • #16
Tom Mattson said:
I have no idea of what you're thinking unless you show how you got those numbers (both are wrong).

Im not sure how to do it because one is in series and one is in parallel.
 
  • #17
So to work out the current i need the p.d and the resistance.
I know the resistance of PQ is 5? I just need to work out the p.d?
 
  • #18
Tom Mattson said:
Because that would mean that the potential difference across QR is zero, which means that no current flows through them. But that is impossible because of conservation of charge at junction Q. Part of the 14V is across PQ and part of it is across QR.

So for part b tell me this: What's the equivalent resistance of the entire network?

7? wild guess.
 
  • #19
Yes, it's 2 ohms. Why are you guessing? Weren't you taught how to find the equivalent resistance of two resistors in series?

Now that you have the total resistance of and the potential difference across the network, use Ohm's law to determine the total current passing through the network.
 
  • #20
Tom Mattson said:
Yes, it's 2 ohms. Why are you guessing? Weren't you taught how to find the equivalent resistance of two resistors in series?

Now that you have the total resistance of and the potential difference across the network, use Ohm's law to determine the total current passing through the network.

No i haven't been taught it. I found something on a website about it.
So i just use the p.d the questions gave me. 14 volts?

So 14/7 = 2A?
 
  • #22
Thank you so much, you have been a massive help. I think i understand it a lot more and i will check that website out. I was used examstutor because our school signed us up to it.

Do you have any ideas about this?
a ) How would the resistance of a piece of wire change if

i) the length were doubled? stay the same? probably not.
ii) the diameter were doubled? would it get less?
 
  • #23
Use the website I gave you. Everything you need to answer that question is there.
 
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