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 2 of 3.
Satish Kumar said:
1 decade ago
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.
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.
Mohan said:
1 decade ago
@sathish
Explain the last step (ie., current in 2kohm = (1*18)/(1+2)=6mA.)
What formula you have use for substitution ?
Explain the last step (ie., current in 2kohm = (1*18)/(1+2)=6mA.)
What formula you have use for substitution ?
Suraj said:
1 decade ago
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.
Javed said:
1 decade ago
Both resistors are in parallel so voltage remains same.
I1 at 2k=V/R i.e. 12/2K=6mA
I1 at 2k=V/R i.e. 12/2K=6mA
Armandwish said:
10 years ago
If 2k resistor connected 1st i.e. beside of voltage source then answer will be 18.
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)
Brajesh said:
1 decade ago
@Raju
R = r1*r2/r1+r2 = 2/3 kOhm
v = 12v
So current I = V/R = 12*3/2=18mA.
R = r1*r2/r1+r2 = 2/3 kOhm
v = 12v
So current I = V/R = 12*3/2=18mA.
Lakshmi said:
1 decade ago
I1=v/R1
=12/1k
=12mA
I2=v/R2
=12/2k
=6mA
total current=12+6=18mA
=12/1k
=12mA
I2=v/R2
=12/2k
=6mA
total current=12+6=18mA
Prasanna kumar said:
1 decade ago
There is formula for this: i2 = (r1*I/r1+r2).
I = total current.
I = total current.
JAGGUBAI said:
1 decade ago
V = IR
I = V/R
So V = 12V ,R = 2000 ohms
I = 12/2000 = 0.006 = 6mA
I = V/R
So V = 12V ,R = 2000 ohms
I = 12/2000 = 0.006 = 6mA
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