Question #226069

A wire of length 2l is and resistance 2R is cut into two pieces and

then stretched twice of its previous length. Now first the:

connected in series the equivalent resistance is R, then they are

connected in parallel the equivalent resistance is Ry. The ratio

 R2,:R1, would be



1
Expert's answer
2021-08-16T17:42:51-0400

Initial length (l)=2l

Resistance(R)=2R

Cut in two pieces

R'=R;

R''=R

First series combination

R1=R+RR_1=R'+R''

R1=R+R=2RR_1=R+R=2R

Parralled combination

1R2=1R+1R\frac{1}{R_2}=\frac{1}{R'}+\frac{1}{R''}

1R2=1R+1R\frac{1}{R_2}=\frac{1}{R}+\frac{1}{R}

1R=2R\frac{1}{R}=\frac{2}{R}

R2=R2R_2=\frac{R}{2}

R2R1=R22R=14\frac{R_2}{R_1}=\frac{\frac{R}{2}}{2R}=\frac{1}{4}


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