Electronics - Parallel Circuits - Discussion

Consider two resistors in parallel. Their total resistance is,

(R1 * R2) / (R1 + R2)

We would like to determine the effect of opening R2. The circuit then contains a single resistor, R1 in series. We would then like to determine whether ? is > or < in the below equation,


(R1 * R2) / (R1 + R2) ? R1

Let's multiply both sides of this inequality by (R1 + R2), which yields,

R1 * R2 ? R1 * (R1 + R2)

expanding the right hand side yields,

R1 * R2 ? R1^2 + R1 * R2

Since resistance is always positive, we have that the right hand side is larger. Therefore, we conclude that resistance will increase.

Consider r1 ,r2 are the resistance which are in parallel.

If any one of resistor get open mean the connection became series.

Ex:

r1 = 2kohm.
r2 = 1kohm.

If both are in parallel means Rt = r1*r2/(r1+r2) = 2/3 = 0.667 kohm.

If both are in series means,

Rt = r1+r2.
Rt = 2+1 = 3kohm.

So resistance value increases when one of them get opened.

Suppose if you have more than 2 resistors say r1=r2=r3=2 ohm then the equivalent resistance is around 1.3 ohm.

Say r3 is opened then the equivalent resistance is 1 ohm hence from this the resistance is decreased. If we consider only 2 resistors in parallel then the resistance will be increased.

In parallel resistance the effective resistance value will be less than the smallest valued resistance. For eg, if a 2 ohm and 3 ohm resistance are in parallel effective resistance will be less than 2 ohm. So when one is opened the effective resistance increases.

Let me explain with the example

Let
R1 = 4 & R2 =4
Then
RT = R1* R2/( R1+R2)
=4*4/4+4
=16/8
=2 ohms


If R1 is open then R1= not available
So RT = R2
=4

That means total resistance increases as one of the resistance is opened.

According to the parallel resistance circuit the equivalent resistance is given by RT=R1||R2= (R1*R2)/(R2+R2. That answer will be less compare with when any one resistance open, because the total resistance is equalto RT=R1 or R2 only. So RT is increases.

Case 1: R1&R2 in parallel

RT=(R1XR2)/(R1+R2)

Case 2 : R2 is open circuited

RT= R1

To understand correctly

case 1: R1=1k R2=1k

RT= (1x10^6)/(2x10^3)

RT= 0.5k

case 2: R1=1k

RT=1k

So RT increases with removal of R2

As we know for a particular resistor in series net resistance is greater than a resistance in parallel if we remove resistor of one of them from a parallel circuit the net result will increases as reverse as in series.

If we consider the two resistance in parallel, r1 &r2, So according law R1*R2/ (R1+R2). If one of them is open means R1=NO, So it will be R2 in Series. So total resistance will decrease.

Three identical resistors are connected in series to a battery. If the current of 12 A flows the battery, how much current flows through any one of the resistors?

Can anyone answer this?