Respuesta :
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:
[tex]P = I^2R[/tex]
The current, I, can be found by Ohm's Law:
[tex]V = IR[/tex]
Case 1:
The equivalent resistance in resistors connected in series is:
[tex]R_{eq} = R_1 + R_2[/tex]
[tex]I = V/R = E/(R_1 + R_2)[/tex]
[tex]P_1 = I^2R_{eq} = \frac{E^2}{(R_1 + R_2)^2}(R_1 + R_2) = \frac{E^2}{(R_1 + R_2)}[/tex]
Case 2:
The equivalent resistance in resistors connected in parallel is:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\\R_{eq} = \frac{R_1R_2}{R_1 + R_2}[/tex]
[tex]I = V/R = \frac{E}{(\frac{R_1R_2}{R_1 + R_2})} = \frac{E(R_1 + R_2)}{R_1R_2}[/tex]
[tex]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}[/tex]
Comparison:
If we compare [tex]P_1[/tex] and [tex]P_2[/tex]:
[tex]P_1 = \frac{E^2}{(R_1 + R_2)}, P_2 = \frac{E^2(R_1 + R_2)}{R_1R_2}\\[/tex]
According to these equations, we can conclude that
[tex]P_2 > P_1[/tex]
You can give numbers to resistances to compare them.
The ratio between the parallel to the series connection is
[tex]\frac{P_s}{P_p} = \frac{R_1 * R_2}{(R_1+R_2)^2}[/tex]
Data;
- R1
- R2
- Energy = E
Power Dissipated in Resistor
In the series connection;
[tex]Ps = E^2/Rs[/tex]
In the parallel connection;
[tex]Pp = E^2/Rp[/tex]
To compare how power is dissipated between the two connection, we would take the ratio between the parallel and series connection
[tex]P_s/P_p = R_p/P_s = R_1*R_2/(R_1+R_2)^2[/tex]
The ratio between the parallel to the series connection is
[tex]\frac{P_s}{P_p} = \frac{R_1 * R_2}{(R_1+R_2)^2}[/tex]
Learn more on resistors here;
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