Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 13)
13.
If a 1 k
and a 2 k resistor are parallel-connected across a 12 V supply, how much current is received by the 2 k resistor?
and a 2 k resistor are parallel-connected across a 12 V supply, how much current is received by the 2 k resistor?Discussion:
28 comments Page 1 of 3.
Arjun said:
7 years ago
I = V/R.
I = 12/1+12/2,
In this 12/2 stands for the 2KOhm resistor.
So it is 6mA.
I = 12/1+12/2,
In this 12/2 stands for the 2KOhm resistor.
So it is 6mA.
(1)
Faizan said:
7 years ago
Here using current divider you can easily find.
Since 1KOhm||2KOhm.
therefore Req=2/3 KOhm.
now using Ohm's law.
I=V/R.
we get I=m18A.
So current in 2KOhm = (1*18)/(1+2)=6mA.
Since 1KOhm||2KOhm.
therefore Req=2/3 KOhm.
now using Ohm's law.
I=V/R.
we get I=m18A.
So current in 2KOhm = (1*18)/(1+2)=6mA.
(1)
Kiran.kk said:
10 years ago
I = V/R.
R1 = 1 kohm.
R2 = 2 kohm.
Their four r2 = 12/2 = 6 ma.
R1 = 1 kohm.
R2 = 2 kohm.
Their four r2 = 12/2 = 6 ma.
(1)
Jaideep said:
1 decade ago
Can we use current divider rule here?
Souvik said:
8 years ago
Circuit current = V/R = 12/(2/3) = 18mA Current through the 2k resistor = total current*opposite site resistor/ sum of the two resistors. that is equal to 18*1/(2+1) = 6mA.
Yogash joshi said:
9 years ago
I = I1 + I2.
I = V/R.
1/R = 1/R1 + 1/R2.
1/R = 1 + 1/2 = 3/2, So R = 2/3.
NOW I = V/R SO, I = 12/ (2/3) = 18.
AND I1= V/ R1 = 12/ 1= 12.
SO IN EQUATION 1.
I = I1 + I2.
18 = 12 + I2.
SO, I2 = 18 - 12 = 6mA.
I = V/R.
1/R = 1/R1 + 1/R2.
1/R = 1 + 1/2 = 3/2, So R = 2/3.
NOW I = V/R SO, I = 12/ (2/3) = 18.
AND I1= V/ R1 = 12/ 1= 12.
SO IN EQUATION 1.
I = I1 + I2.
18 = 12 + I2.
SO, I2 = 18 - 12 = 6mA.
Lakshmi prasanna said:
10 years ago
Voltage is same in parallel circuit so directly 12/2k = 6 mA.
Armandwish said:
10 years ago
If 2k resistor connected 1st i.e. beside of voltage source then answer will be 18.
RASHMI said:
1 decade ago
The voltage through both the resistors are same but the current divides.
According to KCL (the algebraic sum of all the individual current is 0). Therefore the current through R2=6mA.
According to KCL (the algebraic sum of all the individual current is 0). Therefore the current through R2=6mA.
S.BalaKumar said:
1 decade ago
We know the parallel circuit voltage is same. The voltage across the 2 K ohm resistor is 12 v.
Apply I = V/R. I = 12/2 kohm = 6 ma.
Apply I = V/R. I = 12/2 kohm = 6 ma.
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