An engineering company is hired to improve the efficiency of a transmission line. In their report, they discuss ways that will allow them to reduce the current by a factor of four. Assuming this is done, by what factor will the power losses be reduced?
6 years ago
Answered By Leonardo F
The key concept here is how the power dissipation (P), electric current (i) and resistance (R) are related. The formula is:
$P=\frac{V^2}{R}$P=V2R
In which V is the potential difference of the circuit. According to first Ohm's law:
$V=R\times i$V=R×i
We will have:
$P=\frac{\left(R\times i\right)^2}{R}$P=(R×i)2R
$P=R\times i^2$P=R×i2
Basically, the power dissipation is the product between the electric resistance and the square of the electric current. Since the engineering company said that the electric current can be reduced by a factor of 4, the power losses (or dissipation) will be reduced by a factor of $4^2$42 , or 16, if the total electric resistance of the circuit is constant.
6 years ago
Answered By Leonardo F
The key concept here is how the power dissipation (P), electric current (i) and resistance (R) are related. The formula is:
$P=\frac{V^2}{R}$P=V2R
In which V is the potential difference of the circuit. According to first Ohm's law:
$V=R\times i$V=R×i
We will have:
$P=\frac{\left(R\times i\right)^2}{R}$P=(R×i)2R
$P=R\times i^2$P=R×i2
Basically, the power dissipation is the product between the electric resistance and the square of the electric current. Since the engineering company said that the electric current can be reduced by a factor of 4, the power losses (or dissipation) will be reduced by a factor of $4^2$42 , or 16, if the total electric resistance of the circuit is constant.
Hence, the answer is 16.