Electronics - Parallel Circuits - Discussion

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.

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

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.

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.

According to KCL:

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

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.

Thanks @Seshendra.