A wire that has resistance R is cut into two equal pieces . the two parts are joined in parallel. what is the resistance of the combination?

Open in App

Solution

Let the length of the wire be 'l' and area of cross-section be 'A' and resistivity of the material be q

Therefore R = $\frac{q l}{A}$

When it is cut then take each of the piece:-

Its resistance = $\frac{q l}{2 A}$ = $\frac{R}{2}$

When joined in parallel Net resistance = $\frac{R}{4}$

Suggest Corrections

Similar questions

Q. A uniform wire of resistance

R

is cut into two equal pieces, and these pieces are joined in parallel. What is the equivalent resistance of the combination?

A wire of uniform thickness with a resistance of 27 Ω is cut into three equal pieces and they are joined in parallel. The resistance of parallel combination is

Q. A copper wire of resistance

R

is cut in to ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance

Q. A uniform conductor of resistance R is cut into

10

equal pieces. Half of them are joined in series and the remaining half of them are connected in parallel. If these two combination are joined in series, then the effective resistance of the pieces is?

A uniform wire with a resistance of $27 Ω$ is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.

Join BYJU'S Learning Program

A wire that has resistance R is cut into two equal pieces . the two parts are joined in parallel. what is the resistance of the combination?

Let the length of the wire be 'l' and area of cross-section be 'A' and resistivity of the material be qTherefore R =qlAWhen it is cut then take each of the piece:-Its resistance = ql2A =R2When joined in parallel Net resistance = R4

Let the length of the wire be 'l' and area of cross-section be 'A' and resistivity of the material be q

Therefore R = $\frac{q l}{A}$

When it is cut then take each of the piece:-

Its resistance = $\frac{q l}{2 A}$ = $\frac{R}{2}$

When joined in parallel Net resistance = $\frac{R}{4}$