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likoan [24]
3 years ago
6

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 resis

tance. How does the power dissipated by the resistors in these two cases compare?
Physics
1 answer:
mamaluj [8]3 years ago
4 0

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

<u>Case 1: </u>

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)}

<u>Case 2:</u>

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}

<u>Comparison: </u>

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.

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