Calculating Power Loss from Voltage Change at 3 Ohms Resistance

AI Thread Summary
To calculate power loss from voltage change at 3 ohms resistance, first determine the current at 12,000V using the formula P=IV, which results in a higher current and thus greater power loss. Next, calculate the current at 50,000V, which will be significantly lower, leading to reduced power loss. The power wasted can be calculated using the formula I^2R for both voltage scenarios. Finally, find the difference in power loss between the two voltage levels to assess the savings. This approach effectively demonstrates the impact of voltage on power loss in electrical systems.
urszula
Messages
5
Reaction score
0

Homework Statement


a power station delivers 520kW of power to a factory through wires of total resistance of 3 ohms ,how much less power is wasted if the electricity is deliverd at 50,000V rather than 12,000V


Homework Equations


P=IV=I^R =V^/R


The Attempt at a Solution



I tried to use voltage and resistance to calculate power and compare that with 520kW ,but it did not make any sense.Please help me.
 
Physics news on Phys.org
Determine the current flowing at 12000 V and 520 kW. Power wasted I^2R. Next determine the current at 50000 V and 520 kW (it will be much less than earlier). Determine power wasted. Find the difference.
 
Thanks a lot .
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Calculate the individual voltage, current and power in series and para
Replies
1
Views
1K
Power Line Losses: Trivial Problem, Confusing Answer
Replies
9
Views
2K
Voltage Drop in Parallel Receptacle Wiring
Replies
9
Views
2K
Long distance electrical transmission efficiency
Replies
2
Views
2K
Incandescent Bulb - suitable approximation for voltage-current curve
Replies
2
Views
946
Why is power loss = I^2R instead of P=VI or P=I^2R?
Replies
14
Views
3K
How Does Increasing Voltage Reduce Power Loss in Electrical Transmission?
Replies
6
Views
3K
Power Loss, Current, Resistors, and Voltage
Replies
4
Views
2K
Loss of energy in DC and loss reduction in AC circuits
Replies
2
Views
2K
Calculating Power Loss in Transmission Cables
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top