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 1 of 2.
Priyaphilemon said:
1 decade ago
In an AC circuit containing pure inductance the following formula applies:
Current through ac inductance = voltage/opposition to current flow.
I = V/XL.
I = 15/125.
I = 0.12a.
I = 120ma.
Current through ac inductance = voltage/opposition to current flow.
I = V/XL.
I = 15/125.
I = 0.12a.
I = 120ma.
(1)
Ratna said:
8 years ago
Since the voltage source is parllel to both resistances.
(1)
Joseph said:
1 decade ago
Current through inductor = 15/125=0.12 = 120 mA.
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
BISWAJIT oec said:
1 decade ago
After calculate 152MA PUT votage divider rule it is i*R1/R1+R2.
Paban borah(assam) said:
1 decade ago
R1=470.
R2=125.
Circuit current(i)=152mA.
Current through inductor(i2)=(i*R1)/(R1+R2)=120mA.
R2=125.
Circuit current(i)=152mA.
Current through inductor(i2)=(i*R1)/(R1+R2)=120mA.
Darji divyang said:
1 decade ago
Here circuit current is 152mA but dear friends in question said that find the current through an inductor so that answer is 120mA.
15/125=120mA.
15/125=120mA.
Vijay Bairoji 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.
Answer = 152 mA.
But current through inductor is:
IL= (i*R1)/(R1+R2).
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.
Answer = 152 mA.
But current through inductor is:
IL= (i*R1)/(R1+R2).
Sushant said:
1 decade ago
Current through conductor = (152*470)/595.
Prasanta bcet said:
9 years ago
Here, we have to find the source current then put current division rule.
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