Question

20. In a parallel RL circuit, 10 mA flows through the resistor and 4 mA flows through the inductor. What phase angle separates voltage and current in this circuit?
A. 18.6°
B. 28.6°
C. 24.6°
D. 21.8°

Answer 1

Answer:

Option D: 21.8 degrees

Explanation:

In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented  in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.

Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = $\frac{I_l}{I_R} = \frac{x}{y} \frac{4}{10}$, therefore the angle is the arctangent of 4/10:

$arctang(\frac{4}{10} )= 21.801$ degrees.

Answer 2

Answer:

The phase angle between voltage and current in a parallel $RL$ circuit is 21.8°

Explanation:

In this question  input voltage $V_i_n$  across both the components, the resistor and inductor is same since it is  a  parallel $RL$ circuit.  

Current cross resistor $I_R=10 mA$

Current across inductor $I_L=2 mA$  

Inductive reactance $X_L= \frac{ V_i_n}{I_L} = \frac{V_i_n}{(2 mA)}$

Resistance of the resistor $R= \frac{V_i_n}{I_R} = \frac {V_i_n}{(10 mA)}$

Phase angle between voltage and current ∅ $=tan^-^1 R/X_L$

$=tan^-^1 \frac{(V_i_n /10)}{(V_i_n /2)}= tan^-^1 \ \frac{2}{10}=tan^-^1\ 0.4$

=21.8°