An inductor with inductance L = 0.400 H and negligible resistance is connected t
ID: 1616097 • Letter: A
Question
An inductor with inductance L = 0.400 H and negligible resistance is connected to a battery, a switch S , and two resistors, R1 = 5.00 and R2 = 6.00 (Figure 1) . The battery has emf 48.0 V and negligible internal resistance. S is closed at t=0.
G) Apply Kirchhoffs rules to the circuit and obtain a differential equation fori3(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.
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.
I) When i3 has half of its final value, what is i2? Express your answer with the appropriate units.
J) When i3 has half of its final value, what is i1? Express your answer with the appropriate units.
l in R, LExplanation / Answer
Applying kirchhoff's voltage law,
E - i1R1 - i2R2 = 0 ..............(1)
i1 = i2 + i3 ..............(2)
Substituting (2) in (1) we get,
E - i2(R1+R2) - i3R1 = 0 ................(3)
Since R2 and L are in parallel,
i2R2 = L(di3/dt)
=> i2 = (L/R2)(di3/dt) ................(4)
Substituting (4) in (3) we get,
E - (L/R2)(di3/dt)(R1+R2) - i3R1 = 0
=> (di3/dt) = [R2/{L*(R1+R2)}]*(E - i3R1)
=> {di3/(E - i3R1)} = [R2/{L*(R1+R2)}]dt
Integrating both sides,
=> [{ln(E - i3R1)}/R1]0i3 = [R2t/{L*(R1+R2)]0t
=> ln{(E - i3R1)/E} = R1R2t/{L*(R1+R2)
=> (E - i3R1)/E = e[R1R2t/{L*(R1+R2)]
=> i3 = (E/R1)(1 - e[R1R2t/{L*(R1+R2)])