An inductor with inductance L = 0.500 H and negligible resistance is connected to a battery, a switch S, and two resistors, R₁ = 8.00 2 and R₂ = 4.00 2 (Figure 1). The battery has emf 48.0 V and negligible internal resistance. S is closed at t = 0. Figure < 1 of 1 > An inductor with inductance L = 0.5 H and negligible resistance is connected to a battery, a switch S, and two resistors, R₁ = 8.00 2 and R₂ = 4.00 S2 (Figure 1). The battery has emf 48.0 V and negligible internal resistance. S is closed at t=0. Apply Kirchhoffs rules to the circuit and obtain a differential equation for i3 (t). Integrate the equation to obtain an equation for i3 as a function of the time t that has elapsed since S' was closed. Express your answer in terms of the variables R₁, R2, L, E, and t. xa E t= L Submit —| ΑΣΦ Xb Submit 99 Value √x √√x x templates fraction Previous Answers Request Answer Part H Use the equation that you derived in part G to calculate the value of t for which i3 has half of the final value that you calculate in part F Express your answer with the appropriate units. Request Answer Units ? ? X.10n

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Solve Part G and H
An inductor with inductance L = 0.500 H and negligible
resistance is connected to a battery, a switch S, and two
resistors, R₁ = 8.00 2 and R₂ = 4.00 S2 (Figure 1). The
battery has emf 48.0 V and negligible internal resistance.
S is closed at t = 0.
Figure
< 1 of 1
An inductor with inductance L = 0.500 H and negligible
resistance is connected to a battery, a switch S, and two
resistors, R₁ = 8.00 2 and R₂ = 4.00 S2 (Figure 1). The
battery has emf 48.0 V and negligible internal resistance.
S is closed at t=0.
Figure
1 of 1
www
R₁
O
R₂ L
>
Part G
Apply Kirchhoffs rules to the circuit and obtain a differential equation for i3 (t). Integrate the equation to obtain an equation for
i3 as a function of the time t that has elapsed since S was closed.
Express your answer in terms of the variables R₁, R2, L, E, and t.
iz =
[5] ΑΣΦ
xa
E
L
Submit
Xh
•t
√x √x x
μÀ
templates fraction
Previous Answers Request Answer
t= Value
x
Submit Request Answer
Units
Part H
Use the equation that you derived in part G to calculate the value of t for which i3 has half of the final value that you calculated
in part F.
Express your answer with the appropriate units.
IXI
?
?
X 10n
X
Transcribed Image Text:An inductor with inductance L = 0.500 H and negligible resistance is connected to a battery, a switch S, and two resistors, R₁ = 8.00 2 and R₂ = 4.00 S2 (Figure 1). The battery has emf 48.0 V and negligible internal resistance. S is closed at t = 0. Figure < 1 of 1 An inductor with inductance L = 0.500 H and negligible resistance is connected to a battery, a switch S, and two resistors, R₁ = 8.00 2 and R₂ = 4.00 S2 (Figure 1). The battery has emf 48.0 V and negligible internal resistance. S is closed at t=0. Figure 1 of 1 www R₁ O R₂ L > Part G Apply Kirchhoffs rules to the circuit and obtain a differential equation for i3 (t). Integrate the equation to obtain an equation for i3 as a function of the time t that has elapsed since S was closed. Express your answer in terms of the variables R₁, R2, L, E, and t. iz = [5] ΑΣΦ xa E L Submit Xh •t √x √x x μÀ templates fraction Previous Answers Request Answer t= Value x Submit Request Answer Units Part H Use the equation that you derived in part G to calculate the value of t for which i3 has half of the final value that you calculated in part F. Express your answer with the appropriate units. IXI ? ? X 10n X
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