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?
4 mA
6 mA
8 mA
12 mA
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
28 comments Page 2 of 3.

Prasanna kumar said:   1 decade ago
There is formula for this: i2 = (r1*I/r1+r2).

I = total current.

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.

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.

Jaideep said:   1 decade ago
Can we use current divider rule here?

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.

Suchitra said:   1 decade ago
Voltage division formula r2=r1*I/r1+r2;r1=r2*I/r1+r2

JAGGUBAI said:   1 decade ago
V = IR
I = V/R
So V = 12V ,R = 2000 ohms
I = 12/2000 = 0.006 = 6mA

Javed said:   1 decade ago
Both resistors are in parallel so voltage remains same.

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

Clara said:   1 decade ago
V=IR
12 = I(2K)
I=12/2K
I= 6mA

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.


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