Use + to represent the direction of the current shown in the figure and to repre

Question

Use + to represent the direction of the current shown in the figure and to represent the opposite direction. Let epsilon = 45.0V, R_1 = 47.0 Kilo-Ohm, R_2 = 96.0 kilo-Ohm, R_3 = 88.0 kilo- Ohm and L = 15.0 micro- Henry. At times t = 0 i.e., immediately after switch S is closed, what is the value of current i-s? 2.97 mA 2.44mA 6.54 mA 1.97 mA 0.32 mA At time t = 0 i.e., immediately after switch S is closed, what is the value of current i_x? 0.97 mA 0.44 mA 1.54 mA 0.97 mA 1.87 mA For t = infinity (i.e., a long time after the switch is closed)i_t = 1.97 mA 1.44 mA 8.54mA 2.97 mA5.87 mA A long time after switch S has been closed it is suddenly opened. Answer this and the following problem. Immediately at the switch is opened what is the value of i_t 0.97 A 0.44 mA 4.54 mA 9.97 mA 6.87 mA Immediately as the switch is opened, what is the value of i_2? 9.97 mA 2.94 mA 2.54 mA 7.97 mA 3.87 mA

Explanation / Answer

part A :

immediately afte r the swicth is closed

Current across Indcutor = 0

V = I Rnet

I = V/Rnet

I = 45/((47+96)*1000))

I = 0.32 mA option e:

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i1 = V/R1 = 45/47000 = 0.97 mA

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After a suitably long time,

the current reaches its steady state.

The emf across the inductor is zero and we may imagine it being replaced by a wire. The current in R3 is i1 - i2 . Kirchoff's loop rule gives:

i3 = i1 - i2

total current I3 = 45*(47+96)*10^-3/(47*96 + 47*88 + 96*88)

upon solving for i1 and i2   

I1 = 0.44 mA (option b)

I2 = 2.97 mA ( ipiton d)

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D) The left hand branch is now broken .

the current in the branch as zero,

immediately after the switch is openED

i = v/r = 45/(96 + 88) *1000)

i = 0 A
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E) The current in R3 changes less rapidly as it has an inductor in its

branch.

Infact as the switch is opened, it has the same vale as it had

before the switch was opened.

i3 = i1 - i2 = (2.97 -(0.44) mA

i3 = 2.54 mA (option C)

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