Voltage Drop in Parallel Receptacle Wiring

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annamal
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Homework Statement
A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A and a voltage of 110 V, a lamp that contains a 100-W bulb, an overhead light with a 60-W bulb, and various other small devices adding up to 3.00 W. (a) Assuming the power plant that supplies 110 V electricity to the dorm is 10 km away and the two aluminum transmission cables use 0-gauge wire with a diameter of 8.252 mm, estimate the percentage of the total power supplied by the power company that is lost in the transmission.
Relevant Equations
P = ##I^2 R##
V = IR
I know how to solve the question but I am a bit confused about why there is a current of 3 A and a voltage of 110 V requirement for the refrigerator but the power plant only supplies 110 V.

How can there be a voltage drop on the refrigerator load of 110 V, when there is also a voltage drop on the resistance of the wire which this load is in series with?
 
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annamal said:
How can there be a voltage drop on the refrigerator load of 110 V, when there is also a voltage drop on the resistance of the wire which this load is in series with?
It says "the power plant ... supplies 110 V electricity to the dorm", not that it generates 110V.
 
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russ_watters said:
It isn't needed -- it isn't in the Relevant Equation.
But using ## 110 \rm{V} ## is going to be an abominable estimate of the desired measure?

Also, ## R = \rho \frac{L}{A} ## is needed and isn't explicitly listed either as well as some other parameters.
 
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erobz said:
Why isn't the voltage of the transmission lines given in the problem statement I wonder? I think either the voltage at the plant or the voltage at the at the dorm should be given to make a reasonable estimate.
It is not a real life problem.
They just want you to calculate the power needed to overcome that additional resistance, which is in series with the measurable summation load of all the room equipment.

You have the total number of amps flowing in and out the room; therefore, the same amount must flow through that additional resistance (10 km of wires), located between “110 volts” of pushing potential.
 
Lnewqban said:
It is not a real life problem.
They just want you to calculate the power needed to overcome that additional resistance, which is in series with the measurable summation load of all the room equipment.

You have the total number of amps flowing in and out the room; therefore, the same amount must flow through that additional resistance (10 km of wires), located between “110 volts” of pushing potential.
I was thinking it might be a lesson on why transmission lines are high voltage.
 
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haruspex said:
It says "the power plant ... supplies 110 V electricity to the dorm", not that it generates 110V.
Hmm. The appliances each have 110 V across them since the appliances are in parallel with each other?
 
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annamal said:
Hmm. The appliances each have 110 V across them since the appliances are in parallel with each other?
Yep. Its not exactly of consequence here, but that is generally how receptacles are wired. "Ideally" each receptacles is in parallel with the supply. However, they aren't "ideally" wired in parallel ( as in a dedicated set of line to each receptacle from the source ) in reality to save money and construction labor cost.
 
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