What does steady state operation mean for inductor and capacitor?

The implication of assuming steady state condition is that the inductor current will be periodic. That is,

i(t_0+T)=i(t_0)

Inductor current,

i_L(t_0+T)=\frac{1}{L}\int^T_0 v_L (t) dt + i(t_0)

i_L(t_0+T)-i(t_0)=0=\frac{1}{L}\int^T_0 v_L(t) dt

Multiplying both sides by \frac{L}{T}

\frac{1}{T} \int^T_0 v_L(t) dt=0

<v_L(t)>_{avg}=0

As we can see, the LHS is just the average inductor voltage. This is also known as the inductor volt-second balance which states that the average inductor voltage is 0 for periodic inductor current.

Similarly, for a capacitor in steady state, we can write

v_c(t_0+T)=v_c(t_0)

Capacitor voltage,

v_c(t_0+T)=\frac{1}{C} \int^{T+t_0}_{t_0} i_c(t)dt + v_c(t_0)

v_c(t_0+t)-v_c(t_0)=0=\frac{1}{C} \int^{T+t_0}_{t_0} i_c(t)dt

Multiplying both sides by \frac{C}{T}

\frac{1}{T} \int^{T+t_0}_{t_0} i_c(t)dt = 0

<i_c(t)>_{avg} = 0

Thus, under steady state condition, the average capacitor current is 0 over a cycle. This is also known as the capacitor charge balance.

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2 comments on “What does steady state operation mean for inductor and capacitor?

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