• Support PF! Buy your school textbooks, materials and every day products Here!

Please Help Looks simple at first but really found it difficult.

  • Thread starter maobadi
  • Start date
  • #1
22
0
Two resistors Ra and Rb are connected in series with each other between the terminals of a battery giving a constant pd of 49V. When a voltmeter of 500 ohm resistance is connected across each in turn, the reading are given as Va = 14V and Vb = 28V. Determine the resistance of Ra and Rb..
 

Answers and Replies

  • #2
Averagesupernova
Science Advisor
Gold Member
3,554
566
I'll give you this much: Ra has a lower resistance than Rb. You are going to have to show some work here. What have you determined so far? It's basic algebra which I hate to admit I'm a bit rusty at right now. This should be moved to homework section.
 
  • #3
1,758
57
Do you know the formula to find the parallel resistance (Rp) of Ra and the meter (Rm)? If so, can you rearrange it to find Ra if you know Rp and Rm?
 
  • #4
22
0
((500Ra)/(Ra+500))i = 14
((500Rb)/(Rb+500))i = 28

(Ra+Rb)i = 49

This what I have come up with but where do I go from here
 
  • #5
The Electrician
Gold Member
1,246
152
((500Ra)/(Ra+500))i = 14
((500Rb)/(Rb+500))i = 28

(Ra+Rb)i = 49

This what I have come up with but where do I go from here
Your first mistake is in assuming the battery current is the same in the two cases. You need to rewrite your equations as:

((500Ra)/(Ra+500))i1 = 14
((500Rb)/(Rb+500))i2 = 28

You also know that:

((500Ra)/(Ra+500))i1 + Rb*i1 = 49
((500Rb)/(Rb+500))i2 + Ra*i2 = 49

You can eliminate I1 between the the first two equations of each pair, and I2 between the second two of each pair.

This will give you two equations in the two unknowns, Ra and Rb. You should be able to proceed from there.
 
  • #6
1,758
57
Given:
Vs = 49
Va = 14
Vb = 28
Rm = 500 meter resistance

V(Ra) = Vs - Vb
V(Ra) is the voltage across Ra when the meter is across Rb

V(Rb) = Vs - Va

V(Rm) = V(Rp)*V(Ra)/(V(Ra) - V(Rp))
V(Rm) is the voltage across the meter if Ra
were removed and just the meter were in
series with Rb.

Ia = Va(Rm)/Rm
Ia is current through the meter when it is
replacing Ra.

Ib = Vb(Rm)/Rm
Ib is current through the meter when it is
replacing Rb.

Ra = V(Ra)/Ia
Rb = V(Rb)/Ib
 
Last edited:

Related Threads for: Please Help Looks simple at first but really found it difficult.

Replies
16
Views
4K
Replies
10
Views
4K
Replies
5
Views
593
Replies
1
Views
1K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
0
Views
3K
Top