Two copper wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across a potential difference. What would be the ratio of the heat produced in series and in parallel combination? If R_{p} is the total resistance of the wires in parallel connection, why there is a need to write R_{p} as 1/R_{p} while calculating the total resistance of the wire in parallel?

Two Copper wires of the same material, the same length, the same area of cross-section or diameter have the same resistance, say R. If they are connected

a) in series,

The equivalent resistance $R_{s} = R + R = 2R$

Heat produced = $I^2 \times R_{s} \times t= I^2(2R) \times t= 2R \times I^2 \times t$

Effective or equivalent resistance doubles in series circuit

If the two copper wires are connected

b) in parallel

The equivalent resistance is $R_{p}$

$\frac{1}{R_{p}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$

$So R_{p} = \frac{R}{2}$

Effective or equivalent resistance halves in a parallel circuit

Heat = $I^2R_{p} \times t = I^2 \times \frac{R}{2} \times t = \frac{R}{4} \times I^2 \times t $

The ratio of heat = in series/in parallel

= $\frac{2R \times I^2 \times t}{ \frac{R}{4} \times I^2 \times t}$

= 8/1 or 8:1

For parallel connection

$R_{p} = \frac{R \times R}{(R + R)} = \frac{R}{2}$

if you do not want to write

$\frac{1}{R_{p}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$ or $Rp = \frac{R}{2}$

which is the same thing

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