Question

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference

Answer 1

Answer:

when connected in series = Req1=2dl/a

when connected in parallel =Req2= dl/2a.

Req1:Req2=2:1/2=4:1

Answer 2

$\blue\bigstar$Correct Question:-

•Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference .Find the ratio of heat produced in series and parallel combination.

$\red\bigstar$Solution:-

•Suppose the resistance of each one of the two wires is R

The equivalent resistance of the series combination

$\\ \: \: \: \: \sf \pink \star R_s=R+R= \bf 2R$

The heat produced in time t ,

$\: \: \: \longrightarrow \sf H_1= \dfrac{V^{2}}{2R} t \: \: \: \: .....(1)\: \: \: \: \: \bigg( \longmapsto H= \dfrac{V^{2}}{R} t \bigg)$

The equivalent resistance of the parallel combination is

$\\ \: \: \: \: \: \sf R_P= \dfrac{R \times R}{R + R}$

$\\ \sf \: \: \: \: R_p= \dfrac{R^{2}}{2R} = \dfrac{R}{2}$

Heat produced in time t ,

$\\ \sf \: \: \: H^{2}= \dfrac{V^{2}}{ \frac{R}{2} } t = \dfrac{2V^{2}t}{R} \: \: .......(2)$

•Now using equation (1) and (2) we have,

$\\ \: \: \: \: \bullet \: \: \sf\dfrac{H_1}{H_2} = \dfrac{V^{2}t}{2R} \times \dfrac{R}{2V^{2}t}$

$\\ \sf \: \: \: \bullet \: \frac{H_1}{H_2} = \dfrac{1}{4}$

$\\ \: \sf or \star \orange{H_1:H_2} = \blue{1:4} \longmapsto Required \: ratio$

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