Question

Two nichrome wires A and B, each of length 5 cm and of radius 1 cm and 3 cm respectively are connected to each other in series. If a current of 5 A flows through the combination of wires. The ratio of potential difference across wire A to that across wire B is found to be N : 1. Find the value of N.

Answer 1

Pre-requisite Knowledge :-

• Formula for resistance = ρ l/A where ρ ( rho ) is resistivity, l = Length of conductor and A = Area of cross section of conductor.

• Formula for potential difference is given by, V = IR where V is potential difference and I & R are current and Resistance respectively. This is also called ohm's law.

• Area of cross section is given by, πr² where π is a constant and r is the radius of conductor.

Solution :-

Here, in this question we are given that two nichrome wires are connected in series whose length is 5 cm each and the radius of first wire is 1 cm and that of another is 3 cm. Current flowing through the connection is 5 ampere and the ratio of potential difference is N : 1. We are asked to find the value of N.

We will use the concept of ohm's law, according to which :

$\implies\sf\dfrac{V_1}{V_2} =\dfrac {IR_1}{IR_2}$

Here,

• V1 and V2 are the potential difference of wire A and wire B.

$\implies\sf\dfrac{V_1}{V_2} =\dfrac { \not IR_1}{ \not IR_2}$

$\implies\sf\dfrac{V_1}{V_2} =\dfrac { R_1}{R_2}$

Now apply formula of resistance

$\implies\sf\dfrac{V_1}{V_2} =\left[\dfrac{ \dfrac{ \rho \: L_1}{A_1} }{ \dfrac{ \rho L_2}{A_2} }\right]$

Since the material of both wire is same i.e.nichrome, value of rho will be same and also the length of wire is same ( L1 = L2 ), so we get :

$\implies\sf\dfrac{V_1}{V_2} =\left[\dfrac{ \dfrac{ 1}{A_1} }{ \dfrac{ 1}{A_2} }\right]$

$\implies\sf\dfrac{V_1}{V_2} ={ \dfrac{A_2} {A_1}}$

Apply formula of Area

$\implies\sf\dfrac{V_1}{V_2} ={ \dfrac{ \pi(r_2)^{2} } { \pi(r_1)^{2} }}$

$\implies\sf\dfrac{V_1}{V_2} ={ \dfrac{ (r_2)^{2} } {(r_1)^{2} }}$

Substituting value of radius.

$\implies\sf\dfrac{V_1}{V_2} ={ \dfrac{ (3 \: cm)^{2} } {(1 \: cm)^{2} }}$

$\implies\sf\dfrac{V_1}{V_2} ={ \dfrac{ 9 \: cm^{2} } {1 \: cm^{2} }}$

$\implies\sf\dfrac{V_1}{V_2} ={ \dfrac{ 9 } {1 }}$

Since the ratio is 9 : 1, the value of N is 9.

Ans : Value of N is 9

Answer 2

Answer :-

The required value of 'N' is 9 .

Explanation :-

We have two Nichrome wires, A and B.

Resistivity (ρ) of both the wires will be same as they are made up of same material .

Also as the wires are connected in series, so same current (I) will flow through them .

For wire 'A' :-

→ Length of the wire = 5 cm

→ Radius = 1 cm

So, resistance of the wire A (R₁) will be :-

⇒ R₁ = ρL₁/A₁

⇒ R₁ = 5ρ/[π(1)²]

⇒ R₁ = 5ρ/π

Now, according to Ohm's Law potential difference (V₁) across the wire will be :-

⇒ V₁ = IR₁

⇒ V₁ = I(5ρ)/π

⇒ V₁ = 5ρI/π -----(1)

For Wire 'B' :-

→ Length of the wire = 5 cm

→ Radius = 3 cm

So, resistance of wire B (R₂), will be :-

⇒ R₂ = ρL₂/A₂

⇒ R₂ = ρ(5)/[π(3)³]

⇒ R₂ = 5ρ/9π

Again by using Ohm's law, we get potential difference across the wire (V₂) will be :-

⇒ V₂ = IR₂

⇒ V₂ = I(5ρ)/9π

⇒ V₂ = 5ρI/9π -----(2)

______________________________

Putting the values of potential differences of 'A' and 'B' obtained above, we get :-

⇒ V₁ : V₂ = N : 1

⇒ V₁/V₂ = N/1

⇒ 5ρI/π ÷ 5ρI/9π = N/1

⇒ 5ρI/π × 9π/5ρI = N/1

⇒ 9 = N/1

⇒ N = 9