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.
Senthil said:
1 decade ago
My simple way to calculate the current flowing through 2kohm resistance is,
If we are exchange 1KOhm resistance by 2 parallel connected 2KOhm resistance, the total circuit will become 3 parallel connected 2KOhm resistances which are having voltage source of 12v.
Total current flowing through the circuit is 18A. So 6mA current will flow through 3 branches of the circuit.
If we are exchange 1KOhm resistance by 2 parallel connected 2KOhm resistance, the total circuit will become 3 parallel connected 2KOhm resistances which are having voltage source of 12v.
Total current flowing through the circuit is 18A. So 6mA current will flow through 3 branches of the circuit.
Vedanti said:
1 decade ago
I1 = V/R1 = 12mA,
I2 = V/R2 = 6mA,
These are the currents flowing through individual branches. But if the resistances are connected in parallel.
Total current = 18mA.
So in across 1st resistance 12 mA will get drop so 6 mA will flow through 2nd resistor.
I2 = V/R2 = 6mA,
These are the currents flowing through individual branches. But if the resistances are connected in parallel.
Total current = 18mA.
So in across 1st resistance 12 mA will get drop so 6 mA will flow through 2nd resistor.
M.V.KRISHNA/palvoncha said:
1 decade ago
Given R1=1k & R2=2k;
current(I1) at R1=(V/R1); current(I2) at R2=(V/R2)
I1=12/1K; I2=12/2K;
I1=12mA; I2=6mA;
from the question, we need I2. so, option b is right.
current(I1) at R1=(V/R1); current(I2) at R2=(V/R2)
I1=12/1K; I2=12/2K;
I1=12mA; I2=6mA;
from the question, we need I2. so, option b is right.
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.
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.
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.
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)
SEEMA INGOLE said:
1 decade ago
Voltage accros parallel ckt is same so
v=12v
& we want to find out current accros 2kohm resistor so
R=2kohm
as , v=ir
i=v/r
i=12/2
=6
Hence option d is right.
v=12v
& we want to find out current accros 2kohm resistor so
R=2kohm
as , v=ir
i=v/r
i=12/2
=6
Hence option d is right.
HARENDRA KUMAR said:
1 decade ago
In parallel circuit voltage remains same.
So, I1=V/R1=12/1=12mA.
I2=V/R1=12/2=6mA. This is right answer.
So for this question option B is true.
So, I1=V/R1=12/1=12mA.
I2=V/R1=12/2=6mA. This is right answer.
So for this question option B is true.
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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