Why did they use only meter resistance for tau of inductor

[nickmanc86]

Homework Statement

http://imgur.com/uLZdBJC

27. The switch has been closed for
about 1 h. It is then opened at the time defined as t 0 s.
a. Determine the time required for the current iL to drop to
10 μA.
b. Find the voltage VL at t 10μs
c. Calculate VL at t 5tau.

Homework Equations

So I completely worked through this problem. Got Eth (20V) and Rth (1.66MΩ). Used 5H for the inductance and plugged into the formula for IL getting iL = 20V/1.66MΩ * e ^-t/tau where I assumed tau to be 5H/1.66MΩ= 3μs. However the answer comes out incorrect .55μs.

Looking at the solution for the book it calculates tau using only the resistance of the meter 10MΩ and gets a tau of 5μs. This of course gives the same answer as the back of the book. I do not understand why they use only the meters resistance for the tau calculation.

For part B i use the simplified release formula for VLof VL = -Vi * e^-t/tau . I simply took my Vi to be Eth or 20V with my tau of 3μs and solved for VL.

However they proceed to use Vi = I*Rm where Vi = (20/1.66MΩ)*10MΩ =120V giving them VL = 120V * e^-t/tau using their tau of 5μs.

I do not understand why they calculate tau using only Rmeter and then subsequently find Vi using Eth/Rth * Rmeter.

3. My Work
A)
Rth = 2MΩ||10MΩ = 1.66MΩ
Eth = 24V*10MΩ/12MΩ = 20V
Tau = L/R = 5H/1.66MΩ = 3μs

Plugged into
IL = E/R * e^-t/tau -> IL = .55μs
Correct Answer = .92μs

B)
Vi = 20V
Tau = L/R = 5H/1.66MΩ = 3μs

Plugged into
VL = -Vi *e^-t/tau = -0.713V
Correct answer = .81V

 

[berkeman]

Homework Statement

http://imgur.com/uLZdBJC

27. The switch has been closed for
about 1 h. It is then opened at the time defined as t 0 s.
a. Determine the time required for the current iL to drop to
10 μA.
b. Find the voltage VL at t 10μs
c. Calculate VL at t 5tau.

Homework Equations

So I completely worked through this problem. Got Eth (20V) and Rth (1.66MΩ). Used 5H for the inductance and plugged into the formula for IL getting iL = 20V/1.66MΩ * e ^-t/tau where I assumed tau to be 5H/1.66MΩ= 3μs. However the answer comes out incorrect .55μs.

Looking at the solution for the book it calculates tau using only the resistance of the meter 10MΩ and gets a tau of 5μs. This of course gives the same answer as the back of the book. I do not understand why they use only the meters resistance for the tau calculation.

For part B i use the simplified release formula for VLof VL = -Vi * e^-t/tau . I simply took my Vi to be Eth or 20V with my tau of 3μs and solved for VL.

However they proceed to use Vi = I*Rm where Vi = (20/1.66MΩ)*10MΩ =120V giving them VL = 120V * e^-t/tau using their tau of 5μs.

I do not understand why they calculate tau using only Rmeter and then subsequently find Vi using Eth/Rth * Rmeter.

3. My Work
A)
Rth = 2MΩ||10MΩ = 1.66MΩ
Eth = 24V*10MΩ/12MΩ = 20V
Tau = L/R = 5H/1.66MΩ = 3μs

Plugged into
IL = E/R * e^-t/tau -> IL = .55μs
Correct Answer = .92μs

B)
Vi = 20V
Tau = L/R = 5H/1.66MΩ = 3μs

Plugged into
VL = -Vi *e^-t/tau = -0.713V
Correct answer = .81V

Are you sure the problem statement is that the switch has been closed and is then opened at t=0? I could be wrong, but that will result in a *very* large voltage being developed across the inductor, limited only by the parasitic capacitance of the inductor and the circuit (which are not specified).

The problem would make much more sense if the switch were closed at t=0...

 

[nickmanc86]

I double checked the problem in the book and it definitely states the switch was closed for 1h and then opened at a time t = 0s. I suppose it is possible there is a typo in the book (it apparently has many). However, correct or not, they proceed to solve it as stated and end up with those answers. Their initial voltage for part B is 120V which is high relative to the voltage source. Unfortunately out of my realm of understanding (basic circuits FTW).

 

[gneill]

When the switch opens the only available path for the current is via the 10 MΩ meter...

 

[nickmanc86]

When the switch opens the only available path for the current is via the 10 MΩ meter...

Oh my, of course. Wow I feel dumb now. Thank you very much that definitely makes sense.