Electronics - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 15)
15.
The current through R1 will be:


Discussion:
17 comments Page 1 of 2.
Vineeth K said:
9 years ago
48mA is the correct answer.
The total current I eff is V/Reff.
Reff is equivalent resistances of r1 r2 and r3 which is 6/11.
So I=48/6 * 11= 88mA.
88mA is what comes from the source which gets divided into two parts, I1 into resistor r1 and I2 into resistor r2.
Which is given in the diagram through ammeter reading.
V is 48v across all the resistors as the voltage does not get divided in parallel connections.
So, performing I=v/r for each branch r1 r2 and r3 we get a total of 88mA with current through r1 as 48mA, r2 as 24mA and r3 as 16mA.
The total current I eff is V/Reff.
Reff is equivalent resistances of r1 r2 and r3 which is 6/11.
So I=48/6 * 11= 88mA.
88mA is what comes from the source which gets divided into two parts, I1 into resistor r1 and I2 into resistor r2.
Which is given in the diagram through ammeter reading.
V is 48v across all the resistors as the voltage does not get divided in parallel connections.
So, performing I=v/r for each branch r1 r2 and r3 we get a total of 88mA with current through r1 as 48mA, r2 as 24mA and r3 as 16mA.
PRASANNA KUMAR said:
1 decade ago
Here one simple formula is there,
In parallel circuit I = I1+I2.
IN THE CKT SHOWN ONE CURRENT IS LEAVING(0.088A) AND OTHER ENTERING(0.040A) THEN,
I+0.040 = 0.088.
I = 0.088-0.040.
= 88*10^-3 - 40*10^-3 = 48*10^-3 = 48mA.
In parallel circuit I = I1+I2.
IN THE CKT SHOWN ONE CURRENT IS LEAVING(0.088A) AND OTHER ENTERING(0.040A) THEN,
I+0.040 = 0.088.
I = 0.088-0.040.
= 88*10^-3 - 40*10^-3 = 48*10^-3 = 48mA.
Harish kumar said:
1 decade ago
We know voltage across parallel circuits are same.
Given that V = 48v.
By ohm's law,
I=V/R.........(1).
Total resistance R1.R2.R3/R1+R2+R3.
= 1.2.3/1+2+3.
= 1 Ohm.
Substitute R value in 1,
I = 48mA.
Given that V = 48v.
By ohm's law,
I=V/R.........(1).
Total resistance R1.R2.R3/R1+R2+R3.
= 1.2.3/1+2+3.
= 1 Ohm.
Substitute R value in 1,
I = 48mA.
Ragi G R said:
1 decade ago
By simple logic we can find it. The current in the previous branch is 0.088. This current get split into two branches. One of them is 0.040.
So the current in the required branch is 0.088-0.040=48mA
So the current in the required branch is 0.088-0.040=48mA
Vasu said:
1 decade ago
In the answer given by Nirak kumar , Req= 1
but, its wrong. The correct formula in case of 3 r's in || is Req=R1R2R3/(R1R2+R2R3+R3R1).
So here Req will become 6/11.
but, its wrong. The correct formula in case of 3 r's in || is Req=R1R2R3/(R1R2+R2R3+R3R1).
So here Req will become 6/11.
Yamini said:
1 decade ago
The current in the first loop is 88mA,in the second loop is 40mA,
So the current across r1 is the difference between the 2 currents,that is 88m-40m=48mA
So the current across r1 is the difference between the 2 currents,that is 88m-40m=48mA
Dipika said:
1 decade ago
@Niraj is correct answer becoz this three resistsnce are parellel so formula is r1*r2*r3/r1+r2+r3... so total resi r..
& then after ohms law apply...
& then after ohms law apply...
Shiva said:
1 decade ago
I = 0.088.
I1 = ?
I2 = 0.040.
Since I = I1+I2 (Kirchhoff current law).
I1 = I-I2 = 0.088 - 0.040 = 0.048 = 48 x 10^-3 = 48mA.
I1 = ?
I2 = 0.040.
Since I = I1+I2 (Kirchhoff current law).
I1 = I-I2 = 0.088 - 0.040 = 0.048 = 48 x 10^-3 = 48mA.
Azhagusurya said:
1 decade ago
Consider th first node:
Ii = I1+I2.
Ii = 0.088.
& I2 = 0.040.
So I1 = Ii-I2 = 0.088-0.040.
I1 = 0.048 in mA.
I1 = 48mA.
Ii = I1+I2.
Ii = 0.088.
& I2 = 0.040.
So I1 = Ii-I2 = 0.088-0.040.
I1 = 0.048 in mA.
I1 = 48mA.
Seshendra said:
1 decade ago
Voltage across parallel resistors are same.
So voltage across R1 is 48v,
We Know i=v/r.
=48mA.
So voltage across R1 is 48v,
We Know i=v/r.
=48mA.
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