Question

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Question in attachment!!!

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Answer 1

In the above Question ,

[ 1 ] . Resistance between R and S

Here , let us observe the figure carefully.

The resistors RY and YS are in series .

When, resistances are in series , the equivalent resistance is the sum of the resistances.

So ,

=> 1.5 Ohm + 1.5 Ohm

=> 3 OHm .

Now , this is parallel with RS = 1.5 Ohm

For parallel resistances -

Equivalent Resistance = R1 × R 2 / { R 1 + R 2 }

=> { 3 × 1.5 } / 4.5

=> 4.5 / 4.5

=> 1 Ohm ....... A 1

Now,

Resistances in series with XY

=> 1.5 Ohm + 1.5 Ohm + 1 Ohm + 1.5 Ohm + 1.5 Ohm

=> 7 Ohm

This is parallel to ( 1.5 Ohm + 1.5 Ohm = ) the 3 Ohm resistance .

Hence ,

Net resistance between X and Y

=> 7 × 3 / { 7 + 3 }

=> 2.1 Ohm ... A 4

Now , resistances In series CD

=> 2.1 Ohm + 1.5 Ohm + 1.5 Ohm

=> 5.1 Ohm

This is parallel to the ( 1.5 Ohm × 3 = ) 4.5 Ohm resistance .

Equivalent resistance between C and D

=> ( 5.1 × 4.5 ) / ( 9.6 )

=> 2.390625 Ohm ...... A 3

Now , resistances in series , PQ

=> 1.5 Ohm + 1.5 Ohm + 2.390625 Ohm

=> 5.390625 Ohm ...... A2

___________

Answer -

I ) 1 Ohm

2 ) 5.390625 Ohm

3 ) 2.390625 Ohm

4 ) 2.1 Ohm

Answer 2

$\huge\mathfrak\blue{Answer:}$

Concept Used:

• Equivalent resistance in series combination is given by
• R = R1 + R2 ... --------( 1 )
• Equivalent resistance in parallel combination in given by
• R = (R1.R2 )/( R1 + R2 ) ---------( 2 )

Given:

• A number of Resistances arranged as shown in the figure .

To Find:

• Equivalent resistance across branch

1. R and S
2. I and Q
3. C and D
4. X and Y

Solution:

Let the tip of triangle be imagined as point as M

It is clear that Resistance RM and MS are in series combination

Hence using equation (1)

R(RMS) = R(RM) + R(MS).

R(RMS) = ( 1.5 + 1.5 ) ohm

R(RMS) = 3 ohm

Also equivalent of RMS is in parallel with RS

Hence equivalent resistance between R and S

Using equation (2)

R(RS) = ( 3 x 1.5 )/( 3 + 1.5 )

R(RS) = 1 ohm

______________________________

Resistance in series across XY branch

=> ( 1.5 + 1.5 + 1 + 1.5 + 1.5 ) ohm

=> 7 ohm

and

=> ( 1.5 + 1.5 ) ohm

=> 3 ohm

Equivalent resistance across XY is

R(XY) = ( 7 x 3 )/( 7 + 3 )

R(XY) = 21/10

R(XY) = 2.1 ohm

______________________________

Resistance in series across CD branch is

=> ( 1.5 + 2.1 + 1.5 ) ohm

=> 5.1 ohm

And

=> ( 1.5 + 1.5 + 1.5 ) ohm

=> 4.5 ohm

Hence equivalent resistance across C and D

R(CD) = ( 5.1 x 4.5 )/( 5.1 + 4.5 )

R(CD) = 22.9 / 9.6

R(CD) = 2.39 ohm

______________________________

Equivalent Resistance across I and Q is

R(IQ) = ( 1.5 + 2.39 + 1.5 ) ohm

R(IQ) = 5.39 ohm

______________________________

Q1) Resistance across R and S

Answer : 1 ohm

Q2) Resistance across I and Q

Answer : 5.39 ohm

Q3) Resistance across C and D

Answer : 2.39 ohm

Q4) Resistance across X and Y

Answer : 2.1 ohm