We have found in the previous sections that, for all practical purposes, an inductor
can be replaced by a short circuit in a dc circuit after a period of time
greater than five time constants has passed. If in the following circuits
we assume that all of the currents and voltages have reached their final
values, the current through each inductor can be found by replacing
each inductor with a short circuit. For the circuit of
[Fig. 1], for
example,
Fig. 1: Substituting the short-circuit equivalent for the inductor for $t > 5\tau$
$$I_1 = {E \over R_1}= {10V \over 2Ω}=5A$$
For the circuit of
[Fig. 2],
Fig. 2: Establishing the equivalent network for $t > 5\tau$
$$I = {E \over {R_2 || R_3}} = {21V \over 2Ω} = 10.5A$$
Applying the current divider rule,
$$I_1 = {R_3 I \over R_3+R_2} = {(6Ω) (10.5A) \over 6Ω+3Ω}$$
$$I_1 = { 63V \over 9Ω} = 7A$$
In the following examples we will assume that the voltage across
the capacitors and the current through the inductors have reached their
final values. Under these conditions, the inductors can be replaced
with short circuits, and the capacitors can be replaced with open circuits.
Example 1: Find the current $I_L$ and the voltage $V_C$ for the network of Fig. 3.
Fig. 3: For Example 1.
Solution:
$$ I_L = {E \over R_1 + R_2} = {10V \over 5Ω} = 2A$$
$$ V_C = {R_2 E \over R_1 + R_2} = {(3A)(10V) \over 5Ω} = 6V$$
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