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NewtonianAlch

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## Homework Statement

http://img207.imageshack.us/img207/3756/24134024.jpg

N.B: (0+) means

**just after**switch is open, and (0-) means

**just before**switch is open.

I have the solutions for this, but there are some problems I have with understanding how inductors and capacitors work in circuits like this.

1) Before that switch is open

**(t < 0)**, the inductor is essentially a short, and the capacitor is an open circuit

2) When that switch is open, the currennt

**I[itex]_{R}[/itex]**at time (0+) is going to be different to (0-) because resistors unlike capacitors and inductors respond quickly to changes in current.

3) The current through the inductor

**i[itex]_{L}[/itex]**at time (0+) is considered to be the same as (0-) because inductors oppose sudden changes in current.

What I do not understand is when the switch is open, current

**i[itex]_{C}[/itex]**at time (0+) - the current through the capacitor has a value of 0.278mA, I thought the capacitor is acting like an open circuit and like the inductor doesn't respond suddenly, how is there a current flow through it?

The solutions also say

**V[itex]_{C}[/itex](0+)**=

**V[itex]_{C}[/itex](0-)**= 0

I understand that just before the switch is open, the capacitor is acting like a short circuit, and there is no current through it, but how can there be 0 voltage at time (0+) and have a current

**i[itex]^{C}[/itex](0+)**= 0.278mA?

i = Cdv/dt, there is no change in voltage, so current should be zero.

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