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An unknown resistor is connected between the terminals of a 3.00 V battery. Energy is dissipated in the resistor at the rate of 0.745 W. The same resistor is then connected between the terminals of a 1.50 V battery. At what rate is energy now dissipated?

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temdan2001

Answer: P = 0.186W

Explanation: Energy dissipated in this question is tantamount to power.

P = IV

But V = IR

Since the resistor is the same, we can use it in the second solution in the second connection to get power.

Please find the attached file for the solution

Ver imagen temdan2001
Remzwisdom

Answer:

Rate of energy now dissipated(power)=0.186watt(W)

Explanation:

First of all, find the resistance with the givens;

Power(P)=0.745W

Voltage(V)=3,00V

Resistance=R

Using the Power formula;

[tex]P=\frac{V^{2}}{R} \\\\0.745=\frac{3^{2} }{R} \\\\R=\frac{9}{0.745} \\\\R=12.08ohm[/tex]

Now to find the rate of energy currently dissipated with the same resistor at;

R=12.08Ω

V=1.50V

P=p(watt)

[tex]P=\frac{V^{2}}{R} \\\\P=\frac{1.5^{2} }{12.08} \\P=0.186watt(W)[/tex]

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