Electronics - Parallel Circuits - Discussion

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?

[A]. 4 mA
[B]. 6 mA
[C]. 8 mA
[D]. 12 mA

Answer: Option B

Explanation:

No answer description available for this question.

Satish Kumar said: (Dec 7, 2010)  
Since 1KOhm||2KOhm
therefore Req=2/3 KOhm.
now using Ohm's law
I=V/R
we get I=18A.
So current in 2KOhm = (1*18)/(1+2)=6mA.

Raju said: (Mar 26, 2011)  
This 18mA current how? Please explain.

Brajesh said: (Jun 17, 2011)  
@Raju

R = r1*r2/r1+r2 = 2/3 kOhm

v = 12v

So current I = V/R = 12*3/2=18mA.

Mohan said: (Jun 24, 2011)  
@sathish

Explain the last step (ie., current in 2kohm = (1*18)/(1+2)=6mA.)

What formula you have use for substitution ?

Lakshmi said: (Jun 25, 2011)  
I1=v/R1
=12/1k
=12mA

I2=v/R2
=12/2k
=6mA
total current=12+6=18mA

Vineet said: (Jul 3, 2011)  
@Mohan

The formula in last step is voltage divider rule.

Rajnish said: (Oct 30, 2011)  
Not voltage divider but current divider.

Seema Ingole said: (Nov 26, 2011)  
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.

M.V.Krishna/Palvoncha said: (Dec 14, 2011)  
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.

Clara said: (Jan 4, 2012)  
V=IR
12 = I(2K)
I=12/2K
I= 6mA

Javed said: (Jan 19, 2012)  
Both resistors are in parallel so voltage remains same.

I1 at 2k=V/R i.e. 12/2K=6mA

Jaggubai said: (Apr 7, 2012)  
V = IR
I = V/R
So V = 12V ,R = 2000 ohms
I = 12/2000 = 0.006 = 6mA

Suchitra said: (May 8, 2012)  
Voltage division formula r2=r1*I/r1+r2;r1=r2*I/r1+r2

Harendra Kumar said: (Sep 16, 2012)  
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.

Jaideep said: (Nov 14, 2013)  
Can we use current divider rule here?

Vedanti said: (Feb 4, 2014)  
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.

Senthil said: (Feb 22, 2014)  
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.

Prasanna Kumar said: (May 10, 2014)  
There is formula for this: i2 = (r1*I/r1+r2).

I = total current.

Suraj said: (Jan 14, 2015)  
In || circuit the voltage drop is same so v = 12 answer resistance is 2. So effective I = v/R, 12/2 = 6. So answer B.

S.Balakumar said: (Jun 8, 2015)  
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.

Rashmi said: (Jun 24, 2015)  
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.

Armandwish said: (Sep 11, 2015)  
If 2k resistor connected 1st i.e. beside of voltage source then answer will be 18.

Kiran.Kk said: (Jan 17, 2016)  
I = V/R.

R1 = 1 kohm.
R2 = 2 kohm.

Their four r2 = 12/2 = 6 ma.

Lakshmi Prasanna said: (Jan 21, 2016)  
Voltage is same in parallel circuit so directly 12/2k = 6 mA.

Yogash Joshi said: (Aug 20, 2016)  
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.

Souvik said: (Jan 17, 2018)  
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: (Nov 25, 2018)  
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.

Arjun said: (Jan 20, 2019)  
I = V/R.
I = 12/1+12/2,
In this 12/2 stands for the 2KOhm resistor.
So it is 6mA.

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