Electrical Engineering - Inductors - Discussion
Discussion Forum : Inductors - General Questions (Q.No. 17)
17.
An 18 V power supply is connected across a coil with a winding resistance of 180 
. Current flow in the coil is
. Current flow in the coil isDiscussion:
7 comments Page 1 of 1.
Prasad said:
1 decade ago
i=v/r
i=18/180
i=100ma
i=18/180
i=100ma
Surendra choudhary said:
1 decade ago
I = V/R.
I= 18/180= 0.1(option not given).
= 0.1x1000=100 mA.
I= 18/180= 0.1(option not given).
= 0.1x1000=100 mA.
NEERAJ said:
1 decade ago
In dc network inductor coil behave as short circuit. So current (I).
I = V/R.
= 18/180.
= 0.1A or 100mA.
I = V/R.
= 18/180.
= 0.1A or 100mA.
Sandeep yadav said:
1 decade ago
I = V/R.
So 18/180 = 0.1 A or 100 mA.
So 18/180 = 0.1 A or 100 mA.
AMRITA SEN said:
8 years ago
I=V/R.
=18/180,
=0.1A,
1A=1000mA, SO 0.1A=1000*0.1=100mA.
=18/180,
=0.1A,
1A=1000mA, SO 0.1A=1000*0.1=100mA.
Daya Hingmire said:
7 years ago
I= V/R.
= 18/180,
= 0.1A,
= 100mA.
= 18/180,
= 0.1A,
= 100mA.
Bhargav said:
6 years ago
v=ir
so,
i=v/r
by ohms law.
So here we have v=18.
i=?
r=180
then,
i=18/180
=0.1A.
i.e=100mA.
so,
i=v/r
by ohms law.
So here we have v=18.
i=?
r=180
then,
i=18/180
=0.1A.
i.e=100mA.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers