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?

  • 3
    \$\begingroup\$ V0 and Vc are missing from your diagram. \$\endgroup\$
    – JRE
    Commented Jul 9, 2015 at 12:37
  • 2
    \$\begingroup\$ 'finally' means \$t=\infty\$ \$\endgroup\$
    – Chu
    Commented Jul 9, 2015 at 12:46
  • 1
    \$\begingroup\$ Aside from what JRE said, where is the reference node that you will measure \$V_0\$ and \$V_c\$ relative to? \$\endgroup\$
    – The Photon
    Commented Jul 9, 2015 at 16:26

1 Answer 1


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.


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