Question

A WIRE OF RESISTANCE 20 OHM IS BENT TO FORM A CLOSED SQUARE.. WHAT IS THE RESISTANCE ACROSS THE DIAGONAL OF THE SQUARE...??? ​

Answer 1

Answer:

Answer: 5 Ω

Explanation:

In series, equivalent resistance is given by:

R = R₁+R₂+R₃+....

In Parallel, net resistance is given by:

R = ( R₁×R₂×R₃×..)÷(R₁+R₂+R₃+..)

We can think about the wire as connection of 4 pieces having equal resistance(R) and connected in series.

Thus net resistance = 20 Ω = 4 R

⇒R = 5 Ω

Now the wires make a square and we need to find the resistance across the diagonal. Then, two wires are connected in parallel with two other wires.

Resistance of wires connected in series = 2 R = 2 × 5 Ω = 10 Ω

Now, 10 Ω wires connected in parallel with another 10 Ω wire, the net resistance is :

R = (10Ω×10Ω)÷(10Ω+10Ω) = 5Ω

Answer 2

the wire is bent to form a square,then the resistance of the side of the square=20/4=5

Across the diagonal,

first calculate the resistance through the two sides that is

R1=5+5=10

then through the diagonal,R2=root over of(5^2+5^2)=5*root2

as R1 and R2 are parallel to each other,

1/R=1/R1+1/R2

1/R=1/10+1/5*root2

1/R=(root2+root2)/10*root2

1/R=2*root2/10*root2

root2 cancelled in both,hence

1/R=2/10

1/R=1/5

R=5