Answer:
For an additional 130-Ω electrical device, the power used will change by a factor of 0.77.
Explanation:
Power is given as;
P = I²R
Where;
P is the power used due to resistance of the electrical appliance
I is the current passing through the appliance
R is the resistance offered by the appliance
For 70-Ω electrical appliance and 100-Ω electrical appliance, the total power used is;
P = 70I² + 100 I² = (170I²) W
for an additional 130-Ω electrical device, the new the total power used is;
P = 130I² + (170I²) = (300 I²)W
[tex]Change \ in \ power \ used = \frac{I^2(300-170)}{I^2(170)} =0.77[/tex]
Therefore, for an additional 130-Ω electrical device, the power used will change by a factor of 0.77.
