Question

Two identical resistors () are wired in series with a battery of voltage . Suppose one of the resistors is removed from the circuit. How does the power dissipation of the single resistor compare to the power dissipation of that resistor when the second resistor was in the circuit?

Answer 1

Answer:

It's four times larger.

Explanation:

If we have a series circuit, we can get the power dissipated on a resistor R, applying Joule's Law:

P = I²*R

a) When the battery voltage is applied directly to the resistor R, we can apply Ohm's Law to find the current in the circuit, as follows:

$I = \frac{V}{R}$

Replacing this value in the expression for the power dissipated on the resistor R, we get:

$P = \frac{V^{2}}{R^{2} } * R =\frac{V^{2}}{R} (1)$

⇒  $P =\frac{V^{2}}{R}$

When the second resistor is inserted in the circuit, applying Ohm's Law, we can find the new value of  the current I, as follows:

$I = \frac{V}{2*R}$

Replacing this value in (1):

$P = \frac{V^{2}}{4R^{2} } * R =\frac{V^{2}}{4*R} (2)$

We can see that the power dissipation on the resistor R, when the second resistor was in the circuit, according (1) and (2) is 4 times less than the power dissipation of the single resistor, i.e., the power dissipation of the single resistor $(\frac{V^{2}}{R})$  , is four times the power dissipation of that resistor when the second resistor was in the circuit $(\frac{V^{2}}{4*R})$