Power $P$ is to be delivered to a device via transmission cables having resistance $R_C$. If $V$ is the voltage across the device and $I$ is the current through it,find the power wasted and how can it be reduced.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The power wasted in the transmission cables due to their resistance $R_C$ is given by the formula $P_{loss} = I^2 R_C$,where $I$ is the current flowing through the cables.
Since the total power delivered is $P = VI$,we can express the current as $I = P/V$.
Substituting this into the power loss formula,we get $P_{loss} = (P/V)^2 R_C = P^2 R_C / V^2$.
From this expression,it is clear that the power loss is inversely proportional to the square of the voltage $(P_{loss} \propto 1/V^2)$.
Therefore,to reduce the power wasted,the transmission should be done at a very high voltage $V$ and a correspondingly low current $I$.

Explore More

Similar Questions

If $10$ bulbs of $50\,W$ are used for $10$ hours a day,how many units of electricity will be consumed in a month ($30$ days)?

Difficult
View Solution

If the resistance of the filament increases with temperature,what will be the power dissipated in a $220\, V- 100\, W$ lamp when connected to a $110\, V$ power supply?

On giving $220\,V$ to a resistor,the power dissipated is $40\,W$. The value of the resistance is ............... $\Omega$.

How much electric energy is consumed by an average Indian household in one second?

Two electric bulbs whose resistances are in the ratio of $1:2$ are connected in series. The powers dissipated in them have the ratio:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo