An electric power line having a total resista | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Power delivered $P_{out} = 1 \ kW = 1000 \ W$, Voltage $V = 250 \ V$, Resistance $R = 2 \ \Omega$.The current $I$ flowing through the line

Vedclass · Accepted Answer

$96.9$

Vedclass · Answer

(A) Given: Power delivered $P_{out} = 1 \ kW = 1000 \ W$, Voltage $V = 250 \ V$, Resistance $R = 2 \ \Omega$.The current $I$ flowing through the line is $I = \frac{P_{out}}{V} = \frac{1000}{250} = 4 \ A$.The power loss in the transmission line is $P_{loss} = I^2 R = (4)^2 	imes 2 = 16 	imes 2 = 32 \ W$.The total power generated at the source is $P_{in} = P_{out} + P_{loss} = 1000 + 32 = 1032 \ W$.The efficiency $\eta$ is given by $\eta = \left( \frac{P_{out}}{P_{in}} ight) 	imes 100$.$\eta = \left( \frac{1000}{1032} ight) 	imes 100 \approx 96.898 \% \approx 96.9 \%$.

An electric power line having a total resistance of $2 \ \Omega$ delivers $1 \ kW$ of power at $250 \ V$. The percentage efficiency of the transmission line is: (in $\%$)

Similar Questions

$A$ current $i$ passes through a wire of length $l$, radius of cross-section $r$ and resistivity $\rho$. The rate of heat generation is

$A$ battery with $EMF$ $E$ and internal resistance $r$ is connected across a resistance $R$. The power consumption in $R$ will be maximum when:

Two electric bulbs have power ratings $P_1$ and $P_2$ at voltage $V$. They are connected in parallel to a voltage source $V$. What is the total power consumed?

The power wasted in the connecting wires of voltage $V$ and finite resistance $R_C$ from the power station to the device of power $P$ is . . . . . . .

$A$ resistor develops $300 \,J$ of thermal energy in $15 \,s$,when a current of $2 \,A$ is passed through it. If the current increases to $3 \,A$,the energy developed in $10 \,s$ is........$J.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module