Transmission Line Powering Load

Question

Consider a long transmission line (4 km). Its a single conductor line with the inner conductor insulated. The outer shell is steel ground.

Consider two scenarios.

Scenario A: 120 V DC power supply, positive is hooked up to long transmission line, which is connected to 30 Ω load. Negative on the power supply is connected directly to the load.

Scenario B: 120 V DC power supply, positive AND negative are hooked up to long transmission line, which are both connected to the same load in Scenario A.

My question is, will scenario B draw more power or the same power than scenario A? The ground wire is taking a major shortcut in scenario A, so intuitively I would suspect that scenario B would take twice the power to drive the load then scenario A, since the ground wire is adding twice the length of effective wire length.

Excuse me if this is a rookie question, I just can't wrap my mind around it.

Answer 1

The power in the load is $$P_L = I^2 R_L=\left(\frac{V}{R_L + R_{cond} + R_{ground}}\right)^2R_L ,$$ and the power drawn from the supply is $$P_S= IV=\frac{V^2}{R_L + R_{cond} + R_{ground}}$$ For the resistances given, $$P_{LA} = \left(\frac{120}{30+40+0}\right)^2 x 30 = 88.2 W and P_{SA} = 205 W$$ and $$P_{LB} = \left(\frac{120}{30+40+40}\right)^2 x 30 = 35.7 W and P_{SB} = 130 W.$$ In either case, a very inefficient power delivery.

In any case, it's impossible for scenario A to draw twice the supply power unless $$R_L = 0$$ and $$R_{cond}=R_{ground}.$$

Answer 2

In the first case, you have power supply, load, and 4 km of wire.

In the other case, you have power supply, load, 4 km of wire to load and another 4 km of wire back from load.

Of course the scenario with double the wire has more voltage drop and power loss in the wire so current to load is lower.

So because scenario B has more wire and more resistance, it draws less power from supply and load gets less power. It will not draw or need more power.

It might be more apparent when you see your notes that each 4 km of wire has 40 ohm of resistance so power supply sees either 70 or 120 ohms depending on if you have 4 or 8 km of wire.

Answer 3

Power in this case is V^2/R. In your scenario power is smaller in the second case as the R is bigger but V stayed the same.

By R I mean Rload + Rline. But even if you mean power in the load its smaller as well. If you want to calculate the powers separately then ...

Pload = V^2/R *Rload/R.
Pline = v^2/R * Rline/Rload

This is true because the resistors divide the total power just the same as they divide the total voltage. R is still the sum of Rline and Rload

As you can see Pload gets smaller with the square of the total resistance.

Answer 4

My question is, will scenario B draw more power or the same power than scenario A? The ground wire is taking a major shortcut in scenario A, so intuitively I would suspect that scenario B would take twice the power to drive the load then scenario A

it is not stated that scenario A takes a shorter path above. even the picture does not show this,

, since the ground wire is adding twice the length of effective wire length.

length is not everything, real coaxial cables often have lower resistance in the outer conductor. so assuming that the outer and inner conductor have the same resistance in not merited, Also, as it is grounded along its length the ground will also contribute to the outer conductor's conductivity.