Total resistance.. Is my answer correct?

Watch this thread
sarah630
Badges: 10
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 3 months ago
#1
Hii. Please help me confirm whether my answer is correct or not? Thank you. 😀
Attached files
0
reply
Stonebridge
Badges: 13
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 3 months ago
#2
(Original post by sarah630)
Hii. Please help me confirm whether my answer is correct or not? Thank you. 😀
It would help if you could post the original question.
The diagram is a bit ambiguous without a cell or some indication of how the current is meant to flow.
For example, are the resistors supposed to be in parallel groups?
If you have made this question up yourself, I suggest abandoning it and using real questions from a book or past papers.

In short - you cannot solve that circuit question as it stands.

By the way - if the resistors are meant to be in parallel groups, then you have combined them incorrectly.
0
reply
sarah630
Badges: 10
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report Thread starter 3 months ago
#3
(Original post by Stonebridge)
It would help if you could post the original question.
The diagram is a bit ambiguous without a cell or some indication of how the current is meant to flow.
For example, are the resistors supposed to be in parallel groups?
If you have made this question up yourself, I suggest abandoning it and using real questions from a book or past papers.

In short - you cannot solve that circuit question as it stands.

By the way - if the resistors are meant to be in parallel groups, then you have combined them incorrectly.
Here is the original question. I'm supposed to find the current I6 using Thevenin's theorem. So I did the first step of removing the Load resistor, short circuiting the voltage sources and calculating the equivalent resistance.
Attached files
0
reply
Stonebridge
Badges: 13
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 3 months ago
#4
(Original post by sarah630)
Here is the original question. I'm supposed to find the current I6 using Thevenin's theorem. So I did the first step of removing the Load resistor, short circuiting the voltage sources and calculating the equivalent resistance.
That makes it a lot clearer what you are doing now.
I assume R6 was the load resistor and the R values are as you show in your own diagram.
In which case you need to add R1 and R2 in parallel, together with R3, R4 and R5 also in parallel.
On your original diagram you added R1 and R2 in series, for example.
1
reply
sarah630
Badges: 10
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report Thread starter 3 months ago
#5
(Original post by Stonebridge)
That makes it a lot clearer what you are doing now.
I assume R6 was the load resistor and the R values are as you show in your own diagram.
In which case you need to add R1 and R2 in parallel, together with R3, R4 and R5 also in parallel.
On your original diagram you added R1 and R2 in series, for example.
thank you so much!!Now after finding the resistance I am stuck at the mesh equations to find the current. I assume since there are 3 independent loops that means 3 unknown currents? However not sure how to write a mesh equation with 2 voltage sources in the same loop:confused:
0
reply
Stonebridge
Badges: 13
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report 3 months ago
#6
(Original post by sarah630)
thank you so much!!Now after finding the resistance I am stuck at the mesh equations to find the current. I assume since there are 3 independent loops that means 3 unknown currents? However not sure how to write a mesh equation with 2 voltage sources in the same loop:confused:
If I understand your question correctly, the voltages (emfs) in a loop are added 'in series' but taking into account which way they are pointing.
So if you have, say, a clockwise current flowing round the loop, an emf with + pointing clockwise is positive and with + pointing anticlockwise is negative. Then use the single total emf added that way in your equation.
Last edited by Stonebridge; 3 months ago
0
reply
sarah630
Badges: 10
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Report Thread starter 3 months ago
#7
(Original post by Stonebridge)
If I understand your question correctly, the voltages (emfs) in a loop are added 'in series' but taking into account which way they are pointing.
So if you have, say, a clockwise current flowing round the loop, an emf with + pointing clockwise is positive and with + pointing anticlockwise is negative. Then use the single total emf added that way in your equation.
OH I see! Thank you!!!
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

How did your AQA GCSE English Language Paper 1 go?

Loved the paper - Feeling positive (3)
25%
The paper was reasonable (6)
50%
Not feeling great about that exam... (2)
16.67%
It was TERRIBLE (1)
8.33%

Watched Threads

View All