Question

What is the resistance of an air gap?
(ii) Derive an expression for equivalent resistance for the following circuit :
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Answer 1

The resistance of air gap is almost infinity.

Air is a very poor conductor ,there is no free electron present in air so it does not help in the flow of current. From Ohm's law we know that current is inversely proportional to resistance , so the resistance of air is tends to infinity.

The equivalent resistance of the given circuit is {(R2+R3)(R4+R5)+R1(R2+R3+R4+R5)}/(R2+R3+R4+R5)

In the given figure we have 5 resistances , R1,R2,R3,R4,R5.

As, we can see same current flows from the resistance R2 and R3 so we can tell that these two resistances are in series combination

So,

the equivalent resistance of R2 and R3= (R2+R3).

Similarly,

R4 and R5 are also in series combination

So,

the equivalent resistance of R4 and R5 =(R4+R5).

Now ,

two different current is flowing through (R2+R3) and (R4+R5) so these two resistances are connected in parallel combination.

So,

the equivalent resistance will be Rp

1/Rp = 1/(R4+R5)+1/(R2+R3)

or, 1/Rp = (R2+R3+R4+R5)/{(R2+R3)(R4+R5)}

or, Rp = {(R2+R3)(R4+R5)}/(R2+R3+R4+R5)

Now ,

Rp is in series with R1 , as same current is flowing through this.

So,

The final equivalent resistance will be

Req= (R1+Rp)

or, Req=R1+[{(R2+R3)(R4+R5)}/(R2+R3+R4+R5)]

or, Req={(R2+R3)(R4+R5)+R1(R2+R3+R4+R5)}/(R2+R3+R4+R5)

Hence, The equivalent resistance of the given circuit is {(R2+R3)(R4+R5)+R1(R2+R3+R4+R5)}/(R2+R3+R4+R5)

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