Respuesta :
Answer:
The store energy in the inductor is 0.088 J
Explanation:
Given that,
Inductor = 100 mH
Resistance = 6.0 Ω
Voltage = 12 V
Internal resistance = 3.0 Ω
We need to calculate the current
Using ohm's law
[tex]V = IR[/tex]
[tex]I=\dfrac{V}{R+r}[/tex]
Put the value into the formula
[tex]I=\dfrac{12}{6.0+3.0}[/tex]
[tex]I=1.33\ A[/tex]
We need to calculate the store energy in the inductor
[tex]U=\dfrac{1}{2}LI^2[/tex]
[tex]U=\dfrac{1}{2}\times100\times10^{-3}\times(1.33)^2[/tex]
[tex]U=0.088\ J[/tex]
Hence, The store energy in the inductor is 0.088 J
The amount of Energy stored in the inductor is calculated as; 0.088 J
What is the energy stored in the inductor?
We are given;
Inductance; L = 100 mH
Resistance; R = 6.0 Ω
Voltage; V = 12 V
Internal resistance; r = 3.0 Ω
Formula for current with internal resistance is;
I = V/(r + R)
I = 12/(3 + 6)
I = 1.33 A
Formula for energy stored in inductor is;
U = ¹/₂LI²
U = ¹/₂ * 100 * 10⁻³ * 1.33²
U = 0.088 J
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