Electronics - Parallel Circuits - Discussion

Thanks @Seshendra.

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.

According to KCL:

I = I1+I2....I1 = I-I2.

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.

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.

@Harish Kumar is wrong because the total resistance calculation is R1.R2.R3/R1R2+R2R3+R3R1.

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.

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.

@according to vasu as req=6/11
I=V/Req =48*(11/6)=88mA

@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...