Question

A platinum wire of diameter 2.1 mm is connected in series to a 64 µF capacitor and a 0.00188 µH inductor to form an RLC circuit. The resistivity of platinum is 1.1 × 10−7 Ω · m. Calculate the length of the platinum wire when the current is critically damped. Answer in units of m.

Answer 1

As we know that the damping attenuation is defined as R/2L

And in the case of LC circuit -

w (angular frequency) is found to be 1/sqrt(LC)

Critical damping occurs if damping attenuation / w = 1 ie R/2L = w

Now solve the problem step-wise.

Step 1 -

Calculate w = 1/sqrt(64 * 10 ^ -6 * 0.00188 * 10^-6)
= 2.88 * 10 ^ 6 rad/s

Step 2 -

calculate resistance for critical damping
Now as R/2L = w
then R = 2Lw for critical damping
= 2 * 0.00188 * 10 ^ -6 * 2.88 * 10 ^6
=0.0108 ohm

Step 3 -

you want the length of the wire.

For this you need its area
.
R = resistivity * length/ area
so length = R * A / resistivity
And A = pi()* r^2 ( r is the radius of the wire = 0.00105 m)

Therefore -

Length = 0.0108 * pi()*0.00105^2/(1.1*10^-7)

= 0.34 m

So, the length of the platinum wire = 0.34 m (Answer)