I am learning about circuits and I have some questions regarding circuit

given this example:

initial condition: - no initial energy stored before switch is closed - there is current at L1 at t = 0-

I am wondering:

  1. at time t= 0+ what is will be the value of Vc? is it 0V or Vs?
  2. If it is 0V, why? Because it will not satisfy the KVL eqn when switch is closed.
  3. If it is Vs, why so? I understand that Voltage at L can change instatenously but not voltage at C and R. that means at t=0+, should not the VR, C1 and VR2= 0 V? but again this will violates KVL law.
  • \$\begingroup\$ There is an extra R in your line. KVL will be satisfied as long as all voltages in a loop equate back to zero. V2 = x & Vc = 0, just means Vr = -x. \$\endgroup\$
    – Asmyldof
    Mar 28, 2016 at 23:57
  • \$\begingroup\$ Can you clarify please: Is there current in the inductor at t=0- ? \$\endgroup\$
    – berto
    Mar 29, 2016 at 0:31
  • \$\begingroup\$ Voltage across a resistor can change instantaneously. \$\endgroup\$
    – rioraxe
    Mar 29, 2016 at 4:58

1 Answer 1


Use the Laplace transform. Several forms of the voltage across the parallel combination are possible, depending on component values. But this voltage is always zero at t=0

If the system is underdamped, this voltage is an exponentially decaying sine term, hence equal to zero at t=0

If the system is overdamped, the voltage is the difference of two exponentials and, again, zero at t=0

For equal roots, the voltage is of the form \$te^{-at}\$, so equal to zero at t=0

The above assumes that all component values are non-zero, and zero initial conditions. You're question is contradictory on initial conditions - 'no energy storage' and 'there is current in L1 at t=0- '

  • \$\begingroup\$ but isnt having IL doesnot means having v? cause V = di/dt x L. we do not know what di/dt is. I am not really sure but do you mean that Vc(0+) = 0 and to satisfy kvl, all the voltage from voltage source is dissipated at R? \$\endgroup\$ Mar 29, 2016 at 12:48
  • \$\begingroup\$ But DC through a coil is stored energy, so initial conditions would not be zero. It's like having an initial charge on a capacitor. \$\endgroup\$
    – Chu
    Mar 29, 2016 at 14:48

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