Question

Find the equivalent circuit with fewest elements of the following network.

Question 2b. (15 marks) Find the equivalent circuit with the fewest elements of the following network. (15 marks) 6 mH 000 10 uF 14 mH 000 HC 000 35 mH 9 uF 90 uF T

Answer 1

The  
10 μF
and  $90 \ \mu F$ are in series, their effective series capacitance is

* 'S 10 90 10 + 90 C. 9 μF

Now, this  $\\C_s$ is in parallel with the  $9 \ \mu F$ capacitor

therefore, the total capacitance is

$\\C_{total} = C_s + 9 \\\\C_{total} = 9 + 9 \\\\C_{total} = 18 \ \mu C$

The two inductors and are in parallel, their effective parallel inductance is

$\\L_p = \frac{14 * 35}{14 + 35} \\\\L_p = 10 \ mH$

Now this  $L_p$ is in series with the   inductor,

therefore, the total inductance is,

$\\L_{total} = L_p + 6 \\\\L_{total} = 10 + 6 \\\\L_{total} = 16 \ mH$

Therefore, the equivalent circuit with the fewest elements is

L total 16 mol etotal 18 iC