If the voltage across an ohmic resistor is quadrupled, the power expended in the resistor
increases by a factor of 4.
decreases by a factor of 4.
increases by a factor of 16.
none of the preceding
To understand how the power expended in an ohmic resistor changes when the voltage across it is quadrupled, we need to recall the formulas for power and ohm's law.
Ohm's law states that the current through an ohmic resistor is directly proportional to the voltage across it:
V = I * R,
where V is the voltage, I is the current, and R is the resistance of the resistor.
The formula for power in a resistor is given by:
P = I^2 * R = V^2 / R,
where P is the power.
Now, let's consider the scenario where the voltage across the resistor is quadrupled.
If we quadruple the voltage (let's call it V'), then the new voltage will be 4 times the original voltage (4V).
Using Ohm's law, we can rewrite this equation as:
4V = I * R.
If we rearrange this equation to solve for current, we get:
I = (4V) / R.
Now, using the formula for power, we can calculate the new power (P') with the new voltage (4V) and current (I):
P' = [(4V) / R]^2 * R
= (16 * V^2 / R) * R
= 16V^2.
Therefore, when the voltage across an ohmic resistor is quadrupled, the power expended in the resistor increases by a factor of 16.
So, the correct answer is: increases by a factor of 16.