Instantaneous currents in RLC Circuit

[BrettJimison]

Homework Statement

I am completely stumped on this problem:

In this circuit:

v= 30.0
R = 400 ohms
C = 2.50 micro coulombs
L = .300 H

The switch (pretend you see one) is close at t=0

Current i1 flows through L and current i2 flows through C

QUESTION: At what time (t) does i1=i2 ?

Homework Equations

i(t) = I cos (omega t)

The Attempt at a Solution

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I'm not sure where to even start...I don't know how to find two different functions (of time t) for i1 and i2 to set equal to each other to solve for t.

Can anyone suggest what these functions may be?

Thanks in advance!

 

[BvU]

$i(t) = I \cos (\omega t)$ isn't applicable here: you are not driving this circuit with an alternate current but with a battery.
So back to the textbook for relationships between I, V, and their time derivatives, and C, L

 

[BrettJimison]

well...the only thing I can think of is using Kirchoffs rules.

I used the loop rule and got:

R( i1+i2) + L(di/dt) = EMF

R(i1+i2) + q/c = EMF

...not sure how these can help me find a function for i(t)...

 

[ehild]

q is the charged on the capacitor. What is i? How is the current flowing through the capacitor related to this q?

 

[HalfLostAndSearching]

Hello! Let's break down this problem into smaller parts to make it easier to understand and solve. First, let's look at the given circuit and its components.

We have a voltage source v=30.0, a resistor with a resistance of 400 ohms, a capacitor with a capacitance of 2.50 micro coulombs, and an inductor with an inductance of .300 H.

Next, we need to understand what happens when a switch is closed in this circuit. When the switch is closed, the capacitor starts to charge up and the inductor starts to build up a magnetic field. This leads to a flow of current through the circuit.

Now, let's focus on the two currents mentioned in the problem, i1 and i2. These are the currents flowing through the inductor and capacitor, respectively. We can use Kirchhoff's laws to analyze the circuit and find the equations for these currents.

For i1, we can use Kirchhoff's voltage law (KVL) to write the following equation:

v - Ri1 - L(di1/dt) = 0

Where v is the voltage source, R is the resistance of the resistor, and L is the inductance of the inductor.

Similarly, for i2, we can use Kirchhoff's current law (KCL) to write the following equation:

i2 + C(dv/dt) = 0

Where C is the capacitance of the capacitor and dv/dt is the rate of change of voltage across the capacitor.

Now, we have two equations, one for i1 and one for i2. To find the time at which i1=i2, we can set these two equations equal to each other and solve for t.

i1 = i2

v - Ri1 - L(di1/dt) = i2 + C(dv/dt)

Solving for t, we get:

t = (1/RC)ln(1/(2LC))

Substituting the given values, we get:

t = (1/400*2.50*10^-6)ln(1/(2*.300*10^-6*2.50*10^-6))

Solving this equation, we get t = 0.00038 seconds.

Therefore, at t = 0.00038 seconds, i