Electronics - Parallel Circuits - Discussion

Take V = IR relation.

In parallel circuit voltage is constant. So IR must be a constant value if one terminal is open means I = 0. So R became infinity.

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.

We know that if voltage increases, current increases & if resistor increase then current decreases.. then in parallel current ckt as increases total resistance also increases...

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