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1 month ago

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A single resistor is connected across the terminals of a battery. Which one or more of the following changes in voltage and curr

ent leaves unchanged the electric power dissipated in the resistor? (A) Doubling the voltage and reducing the current by a factor of two. (B) Doubling the voltage and increasing the resistance by a factor of four. (C) Doubling the current and reducing the resistance by a factor of four.

1 answer:

Maru [3.3K]1 month ago

7 0

Answer:

Every option provided is accurate

Explanation:

The electrical power dissipated by a single resistor linked to a battery can be expressed as:

$P=VI = I^2 R=\frac{V^2}{R}$

where

V signifies the voltage

I denotes the current

R represents the resistance

Now, let's evaluate each scenario:

A) When the voltage is doubled (V'=2V) while the current is halved (I'=I/2), the resulting power dissipation turns out to be:

$P'=V'I'=(2V)(\frac{I}{2})=VI=P$ --> the power remains the same

B) When the voltage is increased to double (V'=2V) and the resistance quadruples (R'=4R), the new power dissipation becomes:

$P'=\frac{V'^2}{R'}=\frac{(2V)^2}{4R}=\frac{V^2}{R}$ --> the power is unchanged

C) If the current is doubled (I'=2I) while the resistance diminishes to one-fourth (R'=R/4), the new power dissipation is:

$P'=I'^2 R'=(2I)^2(\frac{R}{4})=I^2 R$ --> the power is unchanged

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