Recent content by Meow12

The universe started off as a point and has been expanding for 13.7 billion years. Since this is a finite amount of time, how can the universe be infinite?

I guess ##V_R=V_L=0##. And since ##V_R=iR##, this means ##i=0##? I would've preferred to solve the differential equation, though.

##t\to\infty##

I think it will be ##\mathcal{E}##, the emf of the DC battery.

I guess I could write ##\displaystyle i=\frac{dq}{dt}## to get ##\displaystyle R\frac{dq}{dt}+\frac{q}{C}+L\frac{d^2q}{dt^2}=\mathcal{E}## Then how do I proceed?

How do I solve the differential equation? Please give me a hint.

##\displaystyle 8000\pi L-\frac{1}{8000\pi C}=\pm 13.856## Let us multiply the above equation by 20. We get ##\displaystyle 160000\pi L-\frac{1}{400\pi C}=\pm 13.856\times 20## ----------- (1) The other equation is ##\displaystyle 400\pi L-\frac{1}{400\pi C}=\pm 13.856## ----------- (2)...

Hmm....solving those 2 equations is harder than I expected. The first equation says that ##Z_1=R/0.5=2R## at ##\omega=400\pi## rad/s ##R^2+(X_L-X_C)^2=4R^2## at ##\omega=400\pi## rad/s ##(X_L-X_C)^2=3R^2=192## at ##\omega=400\pi## rad/s ##\displaystyle\left(\omega L-\frac{1}{\omega...

I think I got it! ##R/Z_1=0.5## at ##\omega=2\pi\cdot 200## ##R/Z_2=0.5## at ##\omega=2\pi\cdot 4000## I have 2 equations and 2 unknowns L and C. (R is given.)

Amplifier gain ##A_V## is defined as the ratio of an amplifier's output voltage to its input voltage, i.e. ##\displaystyle\frac{V_R}{V}=\frac{IR}{IZ}=\frac{R}{R}=0.5## at 200 Hz. But this is absurd. Where have I gone wrong? Please nudge me in the right direction.

(a) Substituting the values, I get ##X_L=100\ \Omega##, ##X_C=666.67\ \Omega##. From this, I get ##Z=601\ \Omega##, ##I=49.9\ mA## ##V_R=9.98\ V##, ##V_L=4.99\ V##, ##V_C=33.3\ V## (b) It's possible for the voltage amplitude across the capacitor to be greater than the voltage amplitude across...

I just want to clarify one thing--this is true only when the capacitor is being charged, right? Not when it is being discharged? When the capacitor is being discharged, I think we first meet the - terminal as we traverse the loop in the direction of the assumed current. Because the positive...

No, I didn't mean that. My textbook marks the positive terminal of capacitors in circuits, so I never had to rely on the method you just described. Thanks!

I did not know that. Could you please explain why?

Battery and Capacitor For a battery (or capacitor), the potential increases by ##\mathcal{E}## (or ##\displaystyle\frac{q}{C}##) as we move from -ve to +ve terminal (or plate) regardless of the direction of the assumed current. Resistor and Inductor Suppose we are traversing the loop in the...