Question

Two wires of same metal have the same length but their cross section area are in the ratio 3:2. They are joined in series. If the resistance of the thin wire is 9ohm then the total resistance of the combination will be​

Answer 1

Answer:

Correct option is C

40Ω

Resistance of a wire R=

A

ρl

where ρ= resistivity, l= length, A= cross section of the wire.

As both have same material and length so R∝

A

1

Thus,

R

1

R

2

=

A

2

A

1

=

1

3

⇒R

2

=3R

1

.

here R

1

is the resistance of thicker wire so its resistance R

1

=10Ω (given)

so, R

2

=3(10)=30Ω

As they are connected in series so the equivalent resistance is

R

eq

=R

1

+R

2

=10+30=40Ω

Explanation:

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Answer 2

Given: Resistance of the second thin wire(R2)= 9Ω

            The ratio of their cross-section area= 3:2

To find The total resistance of the combination(R)

Solution: We know that the resistance varies directly with the length(l) and inversely with the cross-section area(A),

                 R∝l/A

Let the resistance of the first thin wire be R1.

The ratio of their cross-section area be A1 and A2

∴A1:A2=3:2 ⇒A1/A2=3/2

So the resistance and the cross-section area for this case can be represented as R1/R2=A2/A1   [∵resistance inversely proportional to the cross-section area ]

⇒R1/9=2/3   [substituting the values of R2, A1, and A2]

⇒R1=(2×9)/3Ω

⇒R1=18/3Ω

⇒R1=6Ω

∴The total resistance of the combination(R)=R1+R2 [∵ R1 and R2 are

                                                                                  joined in series]

⇒R=(6+9)Ω

⇒R=15Ω

Hence the total resistance of the combination(R) is 15Ω.