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?
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.
To determine the factor by which the power emitted by the resistor increases when the voltage is increased, we first need to understand the relationship between voltage, resistance, and power in a DC circuit.
In a DC circuit, ohm's law states that the current flowing through a resistor is equal to the voltage across the resistor divided by the resistance:
I = V / R
Similarly, the power dissipated by a resistor can be calculated using the formula:
P = I^2 * R
Now, let's calculate the power initially emitted by the resistor using the given values:
V1 = 10.0 V (initial voltage)
R = 100 Ω (resistor value)
First, we need to find the current I1:
I1 = V1 / R = 10.0 V / 100 Ω = 0.1 A
Now we can calculate the initial power P1:
P1 = I1^2 * R = (0.1 A)^2 * 100 Ω = 1 W
Next, let's calculate the power emitted when the voltage is increased:
V2 = 20.0 V (new voltage)
Using ohm's law, we find the new current I2:
I2 = V2 / R = 20.0 V / 100 Ω = 0.2 A
Finally, we can calculate the new power P2:
P2 = I2^2 * R = (0.2 A)^2 * 100 Ω = 4 W
To find the factor by which the power emitted by the resistor increases, we divide the new power by the initial power:
Factor = P2 / P1 = 4 W / 1 W = 4
Therefore, the power emitted by the resistor increases by a factor of 4 when the voltage is increased from 10.0 V to 20.0 V.