Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 5)
5.
A 470 
resistor and a coil with 125 inductive reactance are in parallel. Both components are across a 15 V ac voltage source. Current through the inductor is
resistor and a coil with 125 inductive reactance are in parallel. Both components are across a 15 V ac voltage source. Current through the inductor isDiscussion:
13 comments Page 2 of 2.
BISWAJIT oec said:
1 decade ago
After calculate 152MA PUT votage divider rule it is i*R1/R1+R2.
Elavendhan said:
1 decade ago
According Ohms law
V = IR
I = v/R
V = 15 v
I = ?
R1 = 470
R2 = 125
Both the Resistance in parallel so,
R = (R1*R2)/(R1+R2)
= (470*125)/(470+125)
= 98.739
I = 15/98.739
Ans = 152 mA
V = IR
I = v/R
V = 15 v
I = ?
R1 = 470
R2 = 125
Both the Resistance in parallel so,
R = (R1*R2)/(R1+R2)
= (470*125)/(470+125)
= 98.739
I = 15/98.739
Ans = 152 mA
Joseph said:
1 decade ago
Current through inductor = 15/125=0.12 = 120 mA.
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