Electrical Engineering - Series Circuits - Discussion
Discussion Forum : Series Circuits - General Questions (Q.No. 8)
8.
The following resistors (one each) are connected in a series circuit: 470 
, 680 , 1 k, and 1.2 k. The voltage source is 20 V. Current through the 680 resistor is approximately
, 680 , 1 k, and 1.2 k. The voltage source is 20 V. Current through the 680 resistor is approximatelyDiscussion:
23 comments Page 3 of 3.
Rahul said:
1 decade ago
Yes correct.
Nagendra said:
1 decade ago
Current flowing will be same through out the circuit..
So total resistors = 470 + 680 + 1k + 1.2K = 3.35 kohms
i=v/r
20/3.35k
=5.97mA
which is equal to 6mA
So total resistors = 470 + 680 + 1k + 1.2K = 3.35 kohms
i=v/r
20/3.35k
=5.97mA
which is equal to 6mA
GOPALAKRISHNAN said:
1 decade ago
How did you calculate. ?
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