Electronics - Parallel Circuits - Discussion

If any one of the resistor connected in parallel is open can we take that resistor value as 0?

If one of the resistors that are connected in series is opened, then total resistance becomes infinite.

Same question in series then then total resistance what happen?

As circuit is open no current flows & infinite resistance.

In series voltage is constant therefore in the circuit only changes current ane of the fact is that resistance decreases then current also increases as per ohms law.

What will be answered if more than two resistance is connected in a circuit? Like, r1=2, r2=3, r3=4.

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.

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.

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

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.