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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?

Answer :

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

${teks-lihat-gambar} 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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