Question

(i) Derive the relation for parallel combination of resistance with circuit diagram.
(ii) A metallic wire having length(L) and resistance (R) It divides in 7 equal parts, first four
part and next three part are connect I series and this combination is connect in parallel.
Calculate the net resistance in terms of R.
​

Answer 1

Solution :-

(1)

Consider two resistors each of resistance R 1 and R 2 are connected in parallel with each other.

We know that in parallel combination the potential difference across both the resistors

Let , the total current flowing through the circuit be I

Hence ,

➝ I = I 1 + I 2

By ohm's law ,

➝ I = V / R

Hence ,

➝ V / R = V / R 1 + V / R 2

The equivalent resistance when resistors connected in parallel is

➝ 1 / R = 1 / R 1 + 1 / R 2

(2)

Original length = L

The wire is cut into 7 equal parts

New length = L / 7

➝ R ' = ρ L /7A

➝ R ' = ρL/ A x 1 / 7

➝ R ' = R / 7

The first four parts each having resistance R /7 series

➝ Rs = R / 7 + R / 7 + R /7 + R / 7

➝ Rs = 4R /7

The next 3 parts each having resistance R / 7 are connected in series

➝ Rs' = R / 7 + R /7 + R / 7

➝ Rs ' = 3R / 7

Now , Rs and Rs' are connected in parallel

➝ 1 / Rp = 1 / Rs + 1 / Rs '

➝ 1 / Rp = 7 / 4R + 7 /3R

➝ 1 / Rp = 21 + 28 / 12 R

➝ 1 / Rp = 49 / 12 R

➝ Rp = 12R / 49

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The net resistance of them combination is 12 R / 49