Respuesta :
Answer:
The stored energy is 140.7 watt.
The thermal energy is 62.7 watt.
The delivered energy is 203.4 watt.
Explanation:
Given that,
Inductance = 2.8 H
Resistance = 12 Ω
Potential [tex]\epsilon_{0}=89\ V[/tex]
Time = 0.086 s
(a). We need to calculate the energy stored in the magnetic field
Using formula of current
[tex]i=i_{max}(1-e^(\frac{-t}{\tau}))[/tex]
Using formula of energy
[tex]U=\dfrac{1}{2}Li^2[/tex]
On differentiating
[tex]\dfrac{dU}{dt}=Li\frac{di}{dt}[/tex]
[tex]\dfrac{dU}{dt}=L\dfrac{d}{dt}(i_{max}(1-e^(\frac{-t}{\tau}))[/tex]
Again differentiating
[tex]\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})e^{\frac{-t}{\tau}}[/tex]
[tex]\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-\t\times R}{L}})e^{\frac{-t\times R}{L}}[/tex]
Put the value into the formula
[tex]\dfrac{dU}{dt}=\dfrac{(89)^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})e^{\dfrac{-0.086\times12}{2.8}}[/tex]
[tex]\dfrac{dU}{dt}=140.7\ watt[/tex]
(b). We need to calculate the thermal energy
Using formula of thermal energy
[tex]P=i^2R[/tex]
[tex]P=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})^2[/tex]
Put the value into the formula
[tex]P=\dfrac{89^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})^2[/tex]
[tex]P=62.7\ Watt[/tex]
(c). We need to calculate the delivered energy by the battery
Using formula of energy
[tex]P'=P+\dfrac{dU}{dt}[/tex]
[tex]P'=62.7+140.7[/tex]
[tex]P'=203.4\ watt[/tex]
Hence, The stored energy is 140.7 watt.
The thermal energy is 62.7 watt.
The delivered energy is 203.4 watt.
The rate at which energy is stored in the magnetic field is equal to 140.92 Watts.
Given the following data:
Inductance = 2.8 H
Resistance = 12 Ω
Emf = 89 V.
Time = 0.086 seconds.
How to calculate the rate at which energy is stored.
Mathematically, the energy stored in a magnetic field is given by this formula:
[tex]U=\frac{1}{2} LI^2[/tex] ...equation 1.
Also, current is given by:
[tex]I = I_{max}(1-e^{(\frac{-t}{\tau})}} )[/tex]
Differentiating eqn. 1, we have:
[tex]\frac{dU}{dt} =L\frac{d}{dt} (I)\\\\\frac{dU}{dt} =L\frac{d}{dt} ( I_{max}(1-e^{(\frac{-t}{\tau})}} ))\\\\\frac{d^2U}{dt^2} =\frac{\epsilon^2 }{R} (1-e^{\frac{-tR}{L} })e^{\frac{-tR}{L} }[/tex]
Substituting the given parameters into the formula, we have;
[tex]\frac{d^2U}{dt^2} =\frac{89^2 }{12} (1-e^{\frac{-0.086 \times 12}{2.8} })e^{\frac{-0.086 \times 12}{2.8} }\\\\\frac{d^2U}{dt^2} = 660.083(1-e^{(-0.3686)})e^{(-0.3686)\\\\[/tex]
[tex]\frac{d^2U}{dt^2} = 660.083(1-0.6917)0.6917\\\\\frac{d^2U}{dt^2} = 660.083 \times 0.3083 \times 0.6917\\\\\frac{d^2U}{dt^2} = 140.92 \;Watt[/tex]
How to calculate the thermal energy.
The thermal energy that is appearing in the resistance is given by:
[tex]P=\frac{\epsilon^2}{R} (1-e^\frac{-tR}{L} )^2\\\\P=\frac{89^2}{12} (1-e^\frac{-0.086 \times 12}{2.8} )^2\\\\P=660.083(1-e^{(-0.3686)})^2\\\\P=660.083 \times 0.3083^2\\\\P=660.083 \times 0.0951[/tex]
P = 62.77 Watts.
How to calculate the energy delivered by the battery.
[tex]Q=P+\frac{dU}{dt} \\\\Q=62.77 + 140.92[/tex]
Q = 203.69 Watts.
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