Electrical Engineering - Inductors - Discussion

Discussion :: 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

[A]. 1 mA
[B]. 10 mA
[C]. 100 mA
[D]. 1 A

Answer: Option C

Explanation:

No answer description available for this question.

Prasad said: (Mar 4, 2011)  
i=v/r
i=18/180
i=100ma

Surendra Choudhary said: (Apr 4, 2014)  
I = V/R.

I= 18/180= 0.1(option not given).

= 0.1x1000=100 mA.

Neeraj said: (Feb 17, 2015)  
In dc network inductor coil behave as short circuit. So current (I).

I = V/R.
= 18/180.
= 0.1A or 100mA.

Sandeep Yadav said: (May 18, 2015)  
I = V/R.

So 18/180 = 0.1 A or 100 mA.

Amrita Sen said: (Sep 2, 2017)  
I=V/R.
=18/180,
=0.1A,
1A=1000mA, SO 0.1A=1000*0.1=100mA.

Daya Hingmire said: (Mar 4, 2019)  
I= V/R.
= 18/180,
= 0.1A,
= 100mA.

Bhargav said: (Feb 4, 2020)  
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.

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