Please just solve for e and f and show all steps. Thank you!!! Physics II - X3B
ID: 1434126 • Letter: P
Question
Please just solve for e and f and show all steps. Thank you!!! Physics II - X3B - Spring 2016 Offline Problem #7 Due Thursday, March 10, 6:30pm In the circuit shown in the figure, the resistors have resistances R.-R,-10 . the battery has a voltage 6 V, and the inductor is ideal (i.e., it has no internal resistance) with inductance L = 100 mH. The switch has been open for a long time and all currents are initially zero. At time-0 the switch is closed. A "skeleton circuit" showing junctions and loops is also shown below the figure. ho) E /Explanation / Answer
a) Just after switch is closed, current through inductor will be zero.
so current will flow in left side loop only.
b) current through battery, i = E / (R + R) = 6/20 = 0.3 A
PD across inductor, V_L = iR = 0.3 x 10 = 3 Volts
e) Junction Rule equation,
i1 = i2 + iL
and loop rule equation,
e - R1i1 - R2i2 = 0
and in right side loop: R2 i2 - VL = 0
and V_L (t) = Ldi/dt
R2i2 - L (diL / dt) = 0
f) R1 = R2 = R
e - Ri1 - Ri2 = 0
R (i1 + i2) = e
i1 + i2 = e/R
and i1 = i2 + iL
so , i2 + iL + i2 = e/R
i2 = (e/R - iL )/2
putting in right loop side equation,
R(e/R - iL)/2 - L (diL / dt) = 0
L d (iL) / dt = e/2 - R iL/2
di / dt = - (R / 2L)i + (e/2) .......Ans
comparing with di/dt = - lambda i + B
lambda = R / 2L and B = e/2