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 1 of 3.
Kunal said:
5 years ago
Tr=470+680+1000+1200.
Tr=3350.
I=v÷r.
I=20÷3350.
I=0.006.
I=6mA.
Tr=3350.
I=v÷r.
I=20÷3350.
I=0.006.
I=6mA.
(3)
Gurav said:
7 years ago
V / K ohm = mili amp is hat true?
Explain, please .
Explain, please .
Kiran said:
8 years ago
Friends how to calculate 20/3350?
Anthony said:
8 years ago
Current in Series Circuit is constant.
(1)
Prathisha said:
8 years ago
Nice Explanation, Thanks @Amala.
Amala Joseph J said:
8 years ago
@Ashwini S.
We know that, in a series circuit the current flowing through all resistance are same.
So the current flowing through 680-ohm resistor or 1K-ohm or any other resistor connected in a series circuit will be equal to the total current flowing through the circuit.
Hence. if you want to reduce the amount of current flowing through the circuit, then the resistor should not be connected in series. Instead of that you have to connect it in parallel.
This is electrical basics.
We know that, in a series circuit the current flowing through all resistance are same.
So the current flowing through 680-ohm resistor or 1K-ohm or any other resistor connected in a series circuit will be equal to the total current flowing through the circuit.
Hence. if you want to reduce the amount of current flowing through the circuit, then the resistor should not be connected in series. Instead of that you have to connect it in parallel.
This is electrical basics.
(2)
Ashwini S said:
9 years ago
But they have asked to find the current across the 680-ohm resistor, then how 6mA will be the answer?
(3)
ARUN KUMAR M said:
1 decade ago
According to OHMS LAW V = IR.
In a series circuit, current is same, voltage is different.
R = R1+R2+R3+R4 = 3350 V = 20.
I = 20/3350.
I = 5.9 mA.
In a series circuit, current is same, voltage is different.
R = R1+R2+R3+R4 = 3350 V = 20.
I = 20/3350.
I = 5.9 mA.
(5)
H C Chalawadi said:
1 decade ago
Here we have four resistors in series so.
470Ohms+680Ohms+1K (1000 Ohms)+1.2K (1200 Ohms) = 3350 ohms.
Connecting these four resisters in series = Connecting only one 3350 ohms resistor,
V = 20 V.
I = ?
R = 3350 ohms.
I = V/R = 20/3350 = 0.005970149 amps.
Multiply with 1000 as we want result in mA.
So 0.005970149*1000 = 5.970149 mA which is almost equal to 6mA. So Answer is 6mA.
470Ohms+680Ohms+1K (1000 Ohms)+1.2K (1200 Ohms) = 3350 ohms.
Connecting these four resisters in series = Connecting only one 3350 ohms resistor,
V = 20 V.
I = ?
R = 3350 ohms.
I = V/R = 20/3350 = 0.005970149 amps.
Multiply with 1000 as we want result in mA.
So 0.005970149*1000 = 5.970149 mA which is almost equal to 6mA. So Answer is 6mA.
(1)
R.Bairwa said:
1 decade ago
Current will be same in all resistance because of series combination.
(1)
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