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..

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Averagesupernova
Gold Member
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.

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?

((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

The Electrician
Gold Member
((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.

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

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