Question

Consider resistors R1 and R2 connected in series E R1 R2 and in parallel E R1 R2 to a source of emf E that has no internal resistance. How does the power dissipated by the resistors in these two cases compare?

Answer 1

Answer:

The power dissipated by the resistors connected in parallel is higher than that of in series.

Explanation:

Power dissipated by a resistor is given by the following formula:

$P = I^2R$

The current, I, can be found by Ohm's Law:

$V = IR$

Case 1:

The equivalent resistance in resistors connected in series is:

$R_{eq} = R_1 + R_2$

$I = V/R = E/(R_1 + R_2)$

$P_1 = I^2R_{eq} = \frac{E^2}{(R_1 + R_2)^2}(R_1 + R_2) = \frac{E^2}{(R_1 + R_2)}$

Case 2:

The equivalent resistance in resistors connected in parallel is:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\\R_{eq} = \frac{R_1R_2}{R_1 + R_2}$

$I = V/R = \frac{E}{(\frac{R_1R_2}{R_1 + R_2})} = \frac{E(R_1 + R_2)}{R_1R_2}$

$P_2 = I^2R_{eq} = \frac{E^2(R_1 + R_2)^2}{R_1^2R_2^2}\frac{R_1R_2}{(R_1 + R_2)} = \frac{E^2(R_1 + R_2)}{R_1R_2}$

Comparison:

If we compare $P_1$ and $P_2$:

$P_1 = \frac{E^2}{(R_1 + R_2)}, P_2 = \frac{E^2(R_1 + R_2)}{R_1R_2}$

According to these equations, we can conclude that

$P_2 > P_1$

You can give numbers to resistances to compare them.