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
60 mA
30 mA
6 mA
300 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 2 of 3.

RAJAN202509 said:   1 decade ago
In series connection, the current across all the resister is same. So to get the answer. See above solutions.

Anusha said:   1 decade ago
But they asked to calculate current at 680 v right?

Thirumalesh said:   1 decade ago
Equivalent resistance Req = 470+680+1000+1200 = 3350 ohms
"All are connected in series" and
we know that current (I) is same Series connection.

So I=V/Req = 20/3350
I = 6 mA.

Rojalin bhutia said:   1 decade ago
Series current is same, but voltage drop across all the resistors are different.
Here total voltage 20v.

R=680 ohms, 470, 1200, 1000.
Ohms law V=IR.
I = 20/3.35k.
= 5.97 milli amps.

Pushpendra singh said:   1 decade ago
Total resistors =470/1000+680/1000+1+1. 2=3. 35 kohms.
Voltage = 20V.
V=IR.
I=V/R.
=20/3. 35k.
I=5. 97mA.

J khandelwal said:   1 decade ago
Because in series current through all resistor are same.

Sandeep chandraker said:   1 decade ago
Total resistors = 470 + 680 + 1k + 1.2K = 3.35 kohms
V=20v
V=IR
20=I*3.35
I=20/3.35K
I=5.97mA
which is equal to 6mA

KISHORE said:   1 decade ago
ANSWER C IS CORRECT

Srinuivas said:   1 decade ago
In series current is same ,but voltage dropacross all the resistors are different
here total voltage 20v
r=680 ohms ,470,1200,1000
ohms law v=ir
i=20/3.35k
=5.97milliamps

Ramesh said:   1 decade ago
How it is correct?


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