Question

For the circuit shown, find the following: a) v(0+), the voltage across the capacitor right after the switch closes. b) v), the voltage across the capacitor after the switch has been closed for a long time. c) v(T), the voltage across the capacitor after one time constant. 2. 3 S2 I(t) 12 V+ 6 Ω 0.5 F u(t) 3. For the circuit above, write the differential equation for t > 0.

Answer 1

At t = 0 when switch is open V_c - 0(b) therefore V_c (0) = 0V As capacitor doesnot allow sudden always in voltage V (0) = V (o^+) = 0V (b) at t = infinity when switch is closed i = 12/3 + 6 = 12/9 i = 1.33 A therefore V (infinity) = b(1.33) V (infinity) = 8V (c) time constant The time at which voltage across capacitor is 63.2 % of its final value is called time constant therefore V (Tou) = 63.2 % v (infinity) V (Tou) = 0.632 times 8 V (Tou) = 5.056 V 3 i_2 = i (t) = e dv/dt = 0.5 dv/dt Apply KVL to loop 1 -12V + 3i_1 + 6 (i_1-*i_2) = 0 implies -12 V + 3 i_1 + 6 i_1 - 6 i_2 = 0 9 i_1 -6 times 0.5 dv/dt = 12 9 i_1 -3 dv/dt = 12 implies dv/dt = 3i_1 -4