Penta-Inductor Circuit-board

Electricity and Magnetism Level 3

An AC power supply with a voltage ( $V_s$ ) of $30V$ and a frequency ( $f$ ) of $25Hz$ was connected to a circuit of five inductors : $L_1=\frac{17}{\pi}H$ , $L_2=\frac{13}{\pi}H$ , $L_3=\frac{7}{\pi}H$ , $L_4=\frac{19}{\pi}H$ , and $L_5=\frac{4}{\pi}H$ .

Determine the current ( $I$ ) of this circuit in $mA$ .

$Note:$ $X_L = 2 \pi f L_{circuit} \\ I = \frac{V_s}{X_L}$

David's Electricity Set

The answer is 40.

1 solution

David Hontz
Jun 6, 2016

Solution: $L_a = L_1 + L_2 = \frac{17}{\pi} + \frac{13}{\pi} = \frac{30}{\pi}H \\ L_b = L_3+L_4+L_5 = \frac{7}{\pi} + \frac{19}{\pi} + \frac{4}{\pi} = \frac{30}{\pi}H \\ L_{circuit} = \Big( \frac{1}{L_a} + \frac{1}{L_b} \Big) ^{-1} = \Big( \frac{\pi}{30} + \frac{\pi}{30} \Big) ^{-1} = \Big( \frac{\pi}{15} \Big) ^{-1} = \frac{15}{\pi}H$ $X_L = 2 \pi f L_{circuit} = 2 \pi (25Hz)(\frac{15}{\pi}H) = 750Ω$ $I = \frac{30V}{750Ω} = \frac{1}{25}A = 0.04A \Rightarrow I = 40 mA$

Penta-Inductor Circuit-board

David's Electricity Set

The answer is 40.

1 solution

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