If two 2.5 mH inductors are connected in series with a 4.7 $\mathrm { k } \Omega$ resistor and the source voltage is 100 VDC what is the circuit's maximum current?
A) 7.9 mA
B) 21.3 mA
C) 63.2 mA
D) 1.1 mA

Question

If two  2.5 mH  inductors are connected in series with a  4.7 $\mathrm { k } \Omega$  resistor and the source voltage is 100 VDC what is the circuit's maximum current?
A)  7.9 mA
B)  21.3 mA 
C)  63.2 mA
D)  1.1 mA

Quizplus · Accepted Answer

First, we need to calculate the total inductance of the circuit:

Ltotal = L1 + L2 = 2.5 mH + 2.5 mH = 5 mH

Then, we can calculate the angular frequency (w):

w = 2 * pi * f

We don't have the frequency f, but we know that the voltage source is DC, which means that the frequency is 0 Hz. Therefore, w = 0.

Next, we can calculate the total impedance (Z):

Z = R + jwLtotal

j is the imaginary unit (sqrt(-1))

If w = 0, then jwLtotal = 0, and Z = R.

Z = 4.7 kOhm

Finally, we can calculate the maximum current using Ohm's Law:

I = V / Z

I = 100 V / 4.7 kOhm

I = 21.3 mA

Therefore, the answer is B.

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