Question

Ssuppose that Л (capital epsilon) = 58 V, R = 2.40 × 102 Ω, and L = 0.154 H. Initially there is no current in the circuit. Switch S2 is left open, and switch S1 is closed.

Just after S1 is closed, what is the potential differences Vab?

Just after S1 is closed, what is the potential differences Vbc?

A long time (many time constants) after S1 is closed, what is Vab?

A long time (many time constants) after S1 is closed, what is Vbc?

What is Vab at an intermediate time when i = 0.148 A?

What is Vbc at an intermediate time when i = 0.148 A?

Attempt:

So far, I've tried Kirchkoff's Loops Law and some differential equations to try and solve this.

$\Sigma V=0 \\ V_L+V_R+E=0 \\ L\frac{\mathrm{d^2}q}{\mathrm{d}t^2}+R\frac{\mathrm{d}q}{\mathrm{d}t}+E=0$

Then, using the second order linear characteristic equation,

$Lr^2+Rr+E=0$,

I get roots

$r=\frac{-R\pm\sqrt{R^2-4EL}}{2L}$.

Since $R^2>4EL$, I get two real roots,

$q\!\left(t \right )=C_1e^{\frac{-R+\sqrt{R^2-4EL}}{2L}t}+C_2e^{\frac{-R-\sqrt{R^2-4EL}}{2L}t}$.

It gives me that there is no initial current in the circuit after the switch is closed. I can solve for current by taking the derivative. For the sake of saving time, let $\frac{-R+\sqrt{R^2-4EL}}{2L}=r^+$ and $\frac{-R-\sqrt{R^2-4EL}}{2L}=r^-$ . I then get that the current is $q^\prime\!\left(t\right)=C_1r^+e^{r^+t}+C_2r^-e^{r^-t}$ . Since the charge is initially zero, when t is zero, so is i, or q'. Leading to the result that $0=C_1r^++C_2r^-$ . This is as far as I've gotten; I am quite stuck. Any help would be much appreciated as it's due tomorrow at 8 AM....

Answer 1

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