Critically Damped

Let $R'$ be the equivalent resistance. For critical damping:

$1 = \frac{R'}{2} \sqrt{\frac{C}{L}} = \frac{1}{2} \frac{60 R}{ 60 + R} \sqrt{0.0025}$

Solving for $R$ results in $R = 120$ and $R' = 40$ .

Suppose we put $10$ volts on the capacitor and then complete the circuit. The current in the circuit is plotted below for different dampings. For the under-damped case $(R' = 32 )$ , the current oscillates a bit. For the over-damped case $(R' = 48)$ , the current does not oscillate, and instead asymptotically approaches zero after reaching its peak. The critically damped case $(R' = 40)$ is similar to the over-damped case, except that it decays as fast as possible without oscillating.

We have $R_{eq}=\dfrac{60R}{60+R}=2\sqrt {\dfrac{L}{C}}$ . Here $L=4, C=10^{-2}$ . So $\dfrac{60R}{60+R}=2\sqrt {400}=40\implies 3R=120+2R\implies R=\boxed {120}$ Ohm.

Using KVL on this RLC circuit (assuming a potential of $V$ volts across the terminals), we obtain the differential equation:

$V = \frac{60R}{60+R} i(t) + Li'(t) + [v_{C}(0) + \frac{1}{C} \cdot \int_{0}^{t} i(\tau) d\tau]$

which after differentiating with respect to time $t$ yields:

$0 = \frac{60R}{60+R} i'(t) + Li''(t) + \frac{1}{C} i(t)$ (i).

The characteristic equation of (i) has roots equal to: $r = \frac{-\frac{60R}{60+R} \pm \sqrt{(\frac{60R}{60+R})^2 - 4(L)(1/C)}}{2L}$ (ii). If the RLC circuit is to be critically-damped, then we require the discriminant of (ii) to equal zero (i.e. one real root only). Thus:

$(\frac{60R}{60+R})^2 - 4(L)(1/C) = 0 \Rightarrow (\frac{60R}{60+R})^2 = \frac{4L}{C} = \frac{4\cdot4}{.01} = 1600$ (iii)

Solving (iii) for $R$ finally gives $\boxed{R = 120\Omega}.$