Test: Resistance & Inductance in Series - Electrical Engineering (EE) MCQ

XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R²+XL²)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.

 φ=tan^-1(XL/R)=55.1
Since this is an inductive circuit, the current will lag, hence φ= -55.1.

XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R²+XL²)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.
Voltage across resistor= 8.2*7=57.4V.

XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R²+XL²)=12.2ohm.
V=IZ, therefore 100/12.2= 8.2A.
Voltage across resistor= 8.2*7=57.4V.
Voltage across inductor =100-VR= 42.6V.

 XL=2*pi*f*L= 10ohm. Therefore the total impedance =sqrt(R²+XL²)=12.2ohm.
V=IZ, Therefore V=12.2*8.2=100V.

Form the impedance triangle, we get tanφ= XL/R.
Hence φ=tan^-1 (XL/R).

For a series resistance and inductance circuit the phase angle is always a negative value because the current will always lag the voltage.

Form the voltage triangle, we get cosφ= VR/V.
Hence φ=cos^-1VR/V.

An R-L network is a network which consists of a resistor which is connected in series to an inductor.

At DC, the inductor acts as short circuit because the inductive resistance is zero. The frequency of a DC circuit is 0. The inductive resistance=(2*pi*f*L). Therefore, if the frequency is 0, the inductive resistance is zero and it acts as an short circuit.