10 and 11 S9. An R-L series circuit contains a resistor and an inductor. The resistor has a value of6.5 2. The circuit is connected to 120 V and has a current flow of 12 A. What is theinductive reactance of this circuit?#318.01 degrees10. What is the voltage drop across the resistor in the circuit in Question 9?X11. A phase angle meter connected in an R-L series circuit indicates that the current is 28degrees out of phase with the voltage. What is the circuit power factor?20 F312. An R-L series circuit has a power factor of 73%. If the apparent power is 560 kVA, whatis the true power in the circuit?408.8 kvammandED888 FAto t$4RC%5FTV6GFGY44&717HB118

Answered: 9. An R-L series circuit contains a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

10 and 11

S
9. An R-L series circuit contains a resistor and an inductor. The resistor has a value of
6.5 2. The circuit is connected to 120 V and has a current flow of 12 A. What is the
inductive reactance of this circuit?
#3
18.01 degrees
10. What is the voltage drop across the resistor in the circuit in Question 9?
X
11. A phase angle meter connected in an R-L series circuit indicates that the current is 28
degrees out of phase with the voltage. What is the circuit power factor?
20 F3
12. An R-L series circuit has a power factor of 73%. If the apparent power is 560 kVA, what
is the true power in the circuit?
408.8 kva
mmand
E
D
888 FA
to t
$
4
R
C
%
5
F
T
V
6
G
FG
Y
44
&
7
17
H
B
11
8

Expert Solution

Step 1

Question - 10:

The voltage drop across the resistor in an R-L series circuit can be found using Ohm's law, which states that the voltage across a resistor is equal to the current through the resistor multiplied by its resistance:

V = IR

Substituting the values, we get:

V = IR = 12 A * 6.5 ohms = 78 V

So the voltage drop across the resistor in the circuit is 78 V.

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