- 2073 Shrawan
Initial values with switch open to closed (10 Ω, 15 Ω, 2 F)
Short answer
va(0⁻) = 10 V. At t = 0⁺, capacitor voltage remains 10 V; nodal analysis gives va(0⁺) = 50/11 ≈ 4.545 V.
In the network shown in the figure below a steady state is reached with the switch open. At t = 0, the switch is closed. Determine the value of va(0⁻) and va(0⁺). [Figure: 10 V battery (+ at top). The battery positive terminal connects through a 10 Ω resistor to node Va, and through a long 10 Ω resistor along the top wire to the right node Vb. Va connects to Vb through a 15 Ω resistor. A 2 F capacitor connects Vb to the bottom wire. From Va, a switch (closes at t = 0) in series with a 5 Ω resistor connects to the bottom wire.]
Assumptions from the figure: 10 V battery; 10 Ω from the + terminal to node Va; another 10 Ω along the top from the + terminal to Vb; 15 Ω between Va and Vb; C = 2 F from Vb to ground; the switch in series with 5 Ω connects Va to ground and closes at t = 0.
At t = 0⁻ (switch open, steady state)
The capacitor is an open circuit in dc steady state, so no current flows in any resistor. Hence there is no voltage drop and
Va(0⁻) = Vb(0⁻) = 10 V v_C(0⁻) = 10 V
At t = 0⁺
The capacitor voltage cannot change suddenly: Vb(0⁺) = 10 V (the capacitor acts as a 10 V source). The switch is now closed, so the 5 Ω resistor connects Va to ground. KCL at node Va:
(10 − Va)/10 + (Vb − Va)/15 = Va/5
(10 − Va)/10 + (10 − Va)/15 = Va/5
(10 − Va)(1/10 + 1/15) = Va/5
(10 − Va)/6 = Va/5
50 − 5Va = 6Va
Va(0⁺) = 50/11 = 4.545 V
Answer: Va(0⁻) = 10 V and Va(0⁺) = 4.545 V. Note that Va changes suddenly because it is not the capacitor voltage.
✓ Answer checked by Claude Opus 5.5 (high effort)
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