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?
##\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!
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...
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https://deepmind.google/discover/blog/alphageometry-an-olympiad-level-ai-system-for-geometry/...