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 3 of 3.
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.
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.
Rajnish said:
1 decade ago
Not voltage divider but current divider.
Vineet said:
1 decade ago
@Mohan
The formula in last step is voltage divider rule.
The formula in last step is voltage divider rule.
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
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 ?
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.
Raju said:
1 decade ago
This 18mA current how? Please explain.
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