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reaches steady-state at \$t = 0^{-} \$, so I am trying to calculate the initial conditions of \$i\$ (current through inductor) and \$v\$ (voltage across capacitor). It is easy to find that \$i(0) = 10A\$, but I am not able to find \$v(0)\$. What am I missing here?
Just think about how you derived the current of 10 amps. There is a 100 volt supply and there is a 10 ohm resistor - to calculate 10 amps it must mean that all the voltage is across the resistor so, what voltage is across the inductor?
Or, put another way, V = L di/dt and, because the circuit is in steady state (your words not mine), this implies the current is also "steady" so, using that formula, what is the voltage across the inductor?
Actually, the initial state of the capacitor is not defined and it's voltage can be anything somebody might have charged it with.
However, looking at the circuit, we might assume that the switch has been put in the b position before for a long enough time for the circuit to reach steady state.
Now based on the fact that there is a resistor in series with the L and C with the switch in b, we know that the circuit will be 'dead', i.e.: current through L and voltage across C are zero. So let's assume initial state of C having zero voltage.
Now as the voltage over a (ideal) capacitor is always continuous, as is the current through a (ideal) inductor, at \$t=0+\$ the capacitor's voltage will still be zero. Its derivative however is $$u'_C = i_L/C = \frac{10 A}{1/9 F} = 90 V/s $$ Note also that the value of the resistor plays no role in this initial state because the current is determined by the inductor's 10 A.
