Question

In the circuit below, the switch is open for a long time prior to t=0 and closes at t=0. Where $V_A = 5V$, $R_1 = 9 \Omega$, $R_2 = 4 \Omega$, $L = 4H$, $C = 5F$ Write the differential equation for $t \ge 0$ in terms of voltage across the capacitor in the form: $\frac{d^2v_c(t)}{dt^2} + A \frac{dv_c(t)}{dt} + Bv_c(t) = D$ Enter the value of A with three decimal points accuracy

          In the circuit below, the switch is open for a long time prior to t=0 and closes at t=0.
Where $V_A = 5V$, $R_1 = 9 \Omega$, $R_2 = 4 \Omega$, $L = 4H$, $C = 5F$
Write the differential equation for $t \ge 0$ in terms of voltage across the capacitor in the form:
$\frac{d^2v_c(t)}{dt^2} + A \frac{dv_c(t)}{dt} + Bv_c(t) = D$
Enter the value of A with three decimal points accuracy

In the circuit below, the switch is open for a long time prior to t=0 and closes at t=0.
Where VA = 5V, R1 = 9 Ω, R2 = 4 Ω, L = 4H, C = 5F
Write the differential equation for t ≥ 0 in terms of voltage across the capacitor in the form:
(d^2vc(t))/(dt^2) + A (dvc(t))/(dt) + Bvc(t) = D
Enter the value of A with three decimal points accuracy

Added by Jennifer V.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Mahipal Kumawat

Step 1

Before the switch closes, the capacitor is uncharged and can be treated as an open circuit. Therefore, no current flows through the capacitor and the voltage across it is 0V. Show more…

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In the circuit below, the switch is open for a long time prior to t=0 and closes at t=0. Rz MM L i(n) R ve(1) Where V4=5V, R=9, R2=4, L=4 H, C=5 F Write the differential equation for t > 0 in terms of voltage across the capacitor in the form: d^2vc(t)/dt^2 Enter the value of A with three decimal points accuracy.