Electronics - Parallel Circuits - Discussion

The 170 mA is a result of the equivalent resistance of the two parallel resistors. Therefore, if you remove the Resistor R1 then the total current would just be the current flowing through R2 since it is the only resistor left.

IT = (Vs)/(R1||R2) = 170ma,
If R1 is open then,
IT = (Vs)/(R2) = 128ma.

Current always flow through minimum resistance path 170mA.

The Correct option is 170 A because the whole current will pass through R2.

Need resistance for current to flow, if it is open the current cannot flow through the open resistor.

The current flow across the resister depends on the value of the individual resister when they connected in parallel assume r1 as 1 ohm.

And r2 as 1 ohm and and we as 1V than Rt will be 0.5 ohm. It will be 2A, so if r2 removed then It = 1V \r1 = 1v/1ohm = 1A.

So according to ohms law the current flow across r1 is 1A only even though total circuit current is 2A. So when n resisters connected in parallel current flow depends on the value of resister only.

I don't understand the reason the correct answer is 128 mA, since I1 is closed, meaning the total current will flow to I2 which makes it 170 mA. IT = I1+I2 and if I1 is closed IT = I2.

Yes I have same doubt, But when i1 is open then total current will flow through R2 so it should be 170 isn't it ?

But when i1 is open then total current will flow through R2 so it should be 170 isn't it ?

According to kcl law I=I1+I2;.

Whn open i1=0; hence total current is 128mA.