Asked by tyler
A DC circuit has a single battery of voltage V = 10.0 V attached to a resistor R = 100 Ohms. If the voltage is increased to 20.0 V, then by what factor does the power emitted by the resistor increase?
Answers
Answered by
Henry
P2/P1 = ((2V)^2/R)/(V^2/R)=
(4V^2/R)*R/V^2=4V^2/V^2 = 4
Alternate Method:
P1 = V1^2/R = 10^2/100 = 1. Watt.
P2 = V2^2/R = 20^2/100 = 4 Watts.
P2/P1 = 4W/1W = 4
So the power is increased by a factor of
4.
(4V^2/R)*R/V^2=4V^2/V^2 = 4
Alternate Method:
P1 = V1^2/R = 10^2/100 = 1. Watt.
P2 = V2^2/R = 20^2/100 = 4 Watts.
P2/P1 = 4W/1W = 4
So the power is increased by a factor of
4.
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