# Electrical dc shunt resistace question

by oxon88
Tags: electrical, resistace, shunt
 P: 144 1. The problem statement, all variables and given/known data A meter is shunted by a parallel resistance. a) Determine the required value of the shunt resistance if the maximum value of the current I, is 200A. The meter can read a maximum of 1 mA and has a resistance of 0.1 Ω. b) If the shunt is made of copper and has a cross-cectional area of 25cm2 calculate its required length. (for copper take ρ as 1.7x10-8Ωm 2. Relevant equations ohms law (V=I.R) R = ρ.(l/A) 3. The attempt at a solution a) full scale voltage = 0.001 * 0.1 = 0.0001 v shunt resistance Rshunt = 0.0001/200 = 0.5x10-6Ω Rshunt = 0.5 mΩ b) 0.5x10-6 = (1.7x10-8) * ( l / 0.252) (0.5x10-6 / 1.7x10-8) * 0.252 = l l = 1.83824 cm does any of this look right? Thanks.
 Mentor P: 11,689 The cross sectional area of the shunt is already given in cm2. You don't want to square it. Everything else looks okay for the required accuracy.
 P: 144 ok great so is this correct? 0.5x10-6 = (1.7x10-8) * ( l / 0.25) (0.5x10-6 / 1.7x10-8) * 0.25 = l l = 7.353 cm
Mentor
P: 11,689
Electrical dc shunt resistace question

 Quote by oxon88 ok great so is this correct? 0.5x10-6 = (1.7x10-8) * ( l / 0.25) (0.5x10-6 / 1.7x10-8) * 0.25 = l l = 7.353 cm
Looks good.
 P: 1,506 I think you should double check your area units. An area of 25cm^2 is 0.0025m^2. the resistivity is given in ohm.m Are you certain you have been given the cross sectional area as 25cm^2 ? This is a very large area, even for a shunt resistor Also, you have written R = 0.5m.ohms it should be micro.ohms......does not affect your calculation though.
P: 144
 Quote by technician I think you should double check your area units. An area of 25cm^2 is 0.0025m^2. the resistivity is given in ohm.m Are you certain you have been given the cross sectional area as 25cm^2 ? This is a very large area, even for a shunt resistor Also, you have written R = 0.5m.ohms it should be micro.ohms......does not affect your calculation though.
yes the question states a cross sectional area of 25 cm2

Does this look better?

0.5x10-6 = (1.7x10-8) * ( l / 0.0025)

(0.5x10-6 / 1.7x10-8) * 0.0025 = l

l = 0.7353 mm
 P: 1,506 I would say this question does not make sense and is completely unrealistic. A shunt resistor 0.7mm LONG!!!!!!! I would check the source of the question and double double check the numerical information. There is nothing wrong with your method !
Mentor
P: 11,689
 Quote by technician I would say this question does not make sense and is completely unrealistic. A shunt resistor 0.7mm LONG!!!!!!! I would check the source of the question and double double check the numerical information. There is nothing wrong with your method !
I'd say it's plausible: A shunt in the form of a disk clamped in a holder, lots of area for heat dissipation to avoid resistance changes at high currents. Minimal linear expansion for the holder to deal with. To alter the meter's range just swap in the appropriate disk.
 P: 1,506 I still don't believe it! A 0.7mm thick sheet fitted in clamps connected somehow to the terminals of a 1mA movement. Won't the clamps be part of the shunt? Has anyone ever met such a thing? I think there is something wrong with the wording of the original question
Mentor
P: 11,689
 Quote by technician I still don't believe it! A 0.7mm thick sheet fitted in clamps connected somehow to the terminals of a 1mA movement. Won't the clamps be part of the shunt? Has anyone ever met such a thing? I think there is something wrong with the wording of the original question
The clamps only require bulky connections to the cable carrying the 200A. The 1mA to the meter movement (which could be located some distance away from the 'tap') can be carried by lightweight wiring.
 P: 1,506 The calculation should be double checked.... The answer is not 0.7mm There is nothing wrong with your method....check the calculation
 P: 3 Is the answer 7.353 cm or 7.353 m?
 P: 1,506 I got 7.3cm The answers previously given were a mixture of wrong units

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