Two wires of same radius having lengths l1 and l2 and resistivities -1 and -2 are connected in series- The equivalent resistivity will be

The correct option is C ρ_ 1 l _ 1 +ρ_ 2 l _ 2 l _ 1 + l _ 2 The resistance of first wire= R_1=ρ_1l_1A The resistance of second wire= R_2=ρ_2l_ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Resistivity

Two wires of ...

Question

Two wires of same radius having lengths $l_{1}$ and $l_{2}$ and resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity will be

\frac{ρ_{1} l_{2} + ρ_{2} l_{1}}{ρ_{1} ρ_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{l_{1} + l_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{ρ_{1} l_{1} - ρ_{2} l_{2}}{l_{1} - l_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{ρ_{1} l_{2} + ρ_{2} l_{1}}{l_{1} + l_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{l_{1} + l_{2}}$
The resistance of first wire= $R_{1} = \frac{ρ_{1} l_{1}}{A}$
The resistance of second wire= $R_{2} = \frac{ρ_{2} l_{2}}{A}$
Hence the equivalent resistance of circuit= $R_{1} + R_{2} = \frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{A} = \frac{ρ_{e q} (l_{1} + l_{2})}{A}$
$⟹ ρ_{e q} = \frac{ρ_{1} l_{1} + ρ_{2} l_{2}}{l_{1} + l_{2}}$

Suggest Corrections

Similar questions

Two resistors of the same material and cross section area but different lengths $L_{1}$ & $L_{2}$ have their resistances in the ratio

Q. Two spheres

P

and

Q

of equal radii have densities

ρ_{1}

and

ρ_{2}

, respectively. The spheres are connected by a massless string and placed in liquids

L_{1}

and

L_{2}

of densities

σ_{1}

and

σ_{2}

and viscosities

η_{1}

and

η_{2}

, respectively. They float in equilibrium with the sphere

P

L_{1}

and sphere

Q

L_{2}

and the string being taut (see figure). If sphere

P

alone in

L_{2}

has terminal velocity

{\to v}_{P}

and

Q

alone in

L_{1}

has terminal velocity

{\to v}_{Q}

, then

In a circuit two resistances of the same material and cross section area but different lengths $L_{1} a n d L_{2}$ are kept in series. Find the length of the equivalent resistance if it's made of the same material and same cross section area.

Q. Two spheres P and Q of equal radii have densities

ρ_{1}

and

ρ_{2}

, respectively. The spheres are connected by a massless string and placed in liquids

L_{1}

and

L_{2}

of densities

σ_{1}

and

σ_{2}

and viscosities

η_{1}

and

η_{2}

, respectively. They float in equilibrium with the sphere P in

L_{1}

and sphere Q in

L_{2}

and the string being taut. If sphere P along in

L_{2}

has terminal velocity

{\to V}_{P}

and Q along in

L_{1}

has terminal velocity

{\to V}_{Q}

, then

Q. Let

L_{1}

and

L_{2}

be two intersecting lines. lf the image of

L_{1}

w.r.t.

L_{2}

and that of

L_{2}

w.r.t.

L_{1}

concides, then the angle between

L_{1}

and

L_{2}

Join BYJU'S Learning Program

Two wires of same radius having lengths l1 and l2 and resistivities ρ1 and ρ2 are connected in series. The equivalent resistivity will be

The correct option is C ρ1l1+ρ2l2l1+l2The resistance of first wire=R1=ρ1l1AThe resistance of second wire=R2=ρ2l2AHence the equivalent resistance of circuit=R1+R2=ρ1l1+ρ2l2A=ρeq(l1+l2)A⟹ρeq=ρ1l1+ρ2l2l1+l2

Two wires of same radius having lengths $l_{1}$ and $l_{2}$ and resistivities $ρ_{1}$ and $ρ_{2}$ are connected in series. The equivalent resistivity will be