Question

As you can see, I did all calculation, and this is LR and RC circuits lab. I only need this: Present, in text, your interpretation of the data. Tell me what you believe L and C to be in the two measurements.

I did five measurement and calculation. As below. B is value from graph between Voltage(V) in y axis and Time (s) in x axis. The equation is Ae^(-Bt)+y_o

case 1: RC Circuits discharging: the original resistor of R 15 ks2 "charge then discharge Result: B0.00693 Result: B 0.0592 R-10 Ω + 62.5 Ω = 72.5 Ω (because we add the 62.5 Ω in series with the 100 resistor) B 0.0592 16.89s - 144.3s 16.89s 0.001 126F 1 5 × 103Ω t 144.3s = 1.99F R 72.5Ω GasE 2: RC Circuits charging: the original resistor of R-15 k2 "discharge then charge Rcsult: B 0.0590 : = 16.95s B- 0.0590 16.95s R-15x 103_-0.001 129F casc 3: RC Circuits discharging: add a 102 resistor in parallel "charge then discharge Result: B-97.6 and Bea-10 Ω B97.6 0.0102s r 0.0102s C1o0 0.00|02F Case 4: RC Circuits charging: add a 10 Ω resistor in parallel "discharge then charge.. Result B-966 and Ras-10 Ω τ- 0.0 1035s B 96.6 τ 0.0 1035s с 0.00 1035F 10Ω Case 5: LR Circuits charging: bypass the capacitor and put the inductor in there and the 102 resistor still in parallel with the 15kΩ resistor "di "discharg thencharge"

Answer 1

Derive the equation in the given circuit and then match the results would be perfect starting point

For RC circuit

Equation of current in the circuit at any instant t

$V-\frac{q}{C}-iR=0$

dar

dg=0

$\Rightarrow \frac{1}{CV-q}{\mathrm{d} q}=\frac{1}{CR}{\mathrm{d} t}$

Integrating it from t=0 to any time t we get

$q = CV(1-e^{-\frac{t}{CR}})$

$\Rightarrow V = \frac{q}{C}(1+e^{-\frac{t}{CR}})$

$\Rightarrow V = \frac{q}{C}+\frac{q}{C}e^{-\frac{t}{CR}}$

equation you were given

$\Rightarrow V = y_{o}+Ae^{-Bt}$

B=1/CR

C=1/BR

So, your measurements are correct in RC circuit

For the LC circuit

di dt

Similarly

$V=iR+iRe^{-\frac{R}{L}t}$

Here B=R/L

so, L =R/B

But in case 5 you used the wrong equation

So, L=