Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 19)
19.
A 12 V battery is connected across a series combination of 68 
, 47 , 220 , and 33 . The amount of current is
, 47 , 220 , and 33 . The amount of current isDiscussion:
9 comments Page 1 of 1.
Bobby said:
9 years ago
1000 is for, The answer is given in mA and resistance is given in ohms.
So to convert the current from Amp to mA.
So to convert the current from Amp to mA.
Piyush mishra said:
5 years ago
R= R1+R2+R3+R4 = 68+47+220+33 = 368.
i = V/(R) = 12/368 = 0.0326086 In A.
So in milliampere 32.6 mA.
i = V/(R) = 12/368 = 0.0326086 In A.
So in milliampere 32.6 mA.
Deepak pawar said:
1 decade ago
I= V/R.
R= R1+R2+R3+R4.
= 68+47+220+33.
= 368.
I = V/R.
= 12/368*1000.
= 32.6 mA.
R= R1+R2+R3+R4.
= 68+47+220+33.
= 368.
I = V/R.
= 12/368*1000.
= 32.6 mA.
Engr.Raza said:
3 years ago
As in series circuit current remain same so;
I = V/(R1+R2+R3+R3+R4).
I = V/(R1+R2+R3+R3+R4).
S Singh said:
1 decade ago
In S series combination it should be
1/R=1/R1+1/R2+1/R3+1/R4
I=V/R
1/R=1/R1+1/R2+1/R3+1/R4
I=V/R
Devindra said:
9 years ago
Ans is 0.0326A after decimal shifting 32.6mA.
Ashna said:
7 years ago
Please solve this question step by step.
Narender yadav said:
1 decade ago
I = V/(R1+R2+R3+R4)
Guru said:
10 years ago
Why sir 1000?
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