Using voltage divider: [ v_C(0^-) = V_s \times \fracR_2R_1 + R_2 ] [ v_C(0^-) = 12 \times \frac100 , \textk\Omega10 , \textk\Omega + 100 , \textk\Omega ] [ v_C(0^-) = 12 \times \frac100110 = 10.91 , \textV ]
This is a specific request from in the textbook Fundamentals of Electric Circuits (usually by Charles Alexander and Matthew Sadiku, 5th/6th/7th edition). practice problem 7.12 fundamentals of electric circuits
Capacitor voltage cannot change instantly: [ v_C(0^+) = v_C(0^-) = 10.91 , \textV ] For ( t > 0 ), switch is at B. The circuit becomes: capacitor ( C ) in parallel with ( R_2 ), but now ( R_1 ) is disconnected from the source. The only path for discharge is through ( R_2 ). Using voltage divider: [ v_C(0^-) = V_s \times
