I have to find out the value of \$V_{o}\$ and \$V_{c}\$ initially and finally of the this circuit when \$S_{1}\$ and \$S_{2}\$ switches are simultaneously closed.
Edit: \$V_{c}\$ is across the capacitor \$C_{1}\$ and \$V_{o}\$ is across the resistor \$R_{3}\$.
My thought is:
Initially, \$V_{c}=0V\$
Using mesh analysis,
For first loop, \$-25 + I_{1} + 0 +3(I_{1} – I_{2}) – 10 = 0\$ \$or, 4I_{1} – 3I_{2} = 35\$
For second loop, \$ 10 + 3(I_{2} – I_{1}) – 4I_{2} = 0\$ \$or, -3I_{1} + 7I_{2} = -10\$
Solving these two equations, \$I_{1} = 11.31mA\$ and \$I_{2} = 3.42mA\$
Initially, \$V_{o}=4I_{2}\$ \$or, V_{o} = 13.68V\$
By substituting open circuit equivalent for the capacitor,
steady state value for the capacitor is, \$V_{f} = 25V\$
I am confused that the value of \$ t\$ is not given. Then how i can calculate the voltage that is stored in capacitor. Also, how i can calculate the value of \$RC\$. Or I just have to assume that the capacitor is fully charged?
Best Answer
At t=0, Vo is zero too as there is no current flow through R3.
At t=∞, Vo = 10x4/7 V and Vc = 25+40/7 V
Remember that right pin of capacitor is not at zero volts but have some value, calculate by applying voltage divider rule at R3 and R2.