Answer:
9.72 Ω
Explanation:
given,
Potential of battery,V = 9 V
Energy of the resistor = 2.1 x 10⁵ J
time = 7 Hr
= 7 x 3600 = 25200 s
Power,P =[tex]\dfrac{Energy}{time}[/tex]
=[tex]\dfrac{2.1\times 10^5}{25200}[/tex]
= 8.33 W
Resistance=[tex]\dfrac{V^2}{P}[/tex]
=[tex]\dfrac{(9)^2}{8.33}[/tex]
= 9.72 Ω
the resistance of the resistor is equal to 9.72 Ω
The resistance of the resistor connected across the terminals of a 9.0-V battery, which delivers 2.1 x 105 J of energy to the resistor in 7.0 hours will be R=9.72[tex]\Omega[/tex]
It is given that
Volatge =9 volts
Energy = [tex]2.1\times 10^{5}\ J[/tex]
Time = 7 Hours = [tex]7\times 3600=25200\ sec[/tex]
we know that
[tex]Power= \dfrac{Energy}{Time }[/tex]
[tex]Power= \dfrac{ 2.1\times10^{5}}{25200 }=8.33\ W[/tex]
Now we also know that
[tex]P=V^{2}R[/tex]
[tex]R=\dfrac{v^2}{P}[/tex]
[tex]R=\dfrac{9^2}{8.33}=9.72\Omega[/tex]
Thus the resistance of the resistor connected across the terminals of a 9.0-V battery, which delivers 2.1 x 105 J of energy to the resistor in 7.0 hours will be R=9.72[tex]\Omega[/tex]
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