Two wires of equal diameters, of resistivities $p_{1}$ and $p_{2}$ and lengths $l_{1}$ and $l_{2}$ , respectively, are joined in series. The equivalent resistivity of the combination is

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

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\frac{p_{1} l_{2} + p_{2} l_{1}}{l_{1} - l_{2}}

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\frac{p_{1} l_{2} - p_{2} l_{1}}{l_{1} + l_{2}}

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\frac{p_{1} l_{1} - p_{2} l_{2}}{l_{1} + l_{2}}

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Solution

The correct option is A $\frac{p_{1} l_{1} + p_{2} l_{2}}{l_{1} + l_{2}}$
Equal diameter means equal area
In series
$R_{n e t} = R_{1} + R_{1} \frac{p_{n e t} (l_{1} + l_{2})}{A} = \frac{p_{1} + l_{1}}{A} + \frac{p_{2} l_{2}}{A} p_{n e t} = \frac{p_{1} l_{1} + p_{2} l_{2}}{l_{1} + l_{2}}$

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Two wires of equal diameters, of resistivities p1 and p2 and lengths l1 and l2, respectively, are joined in series. The equivalent resistivity of the combination is

The correct option is A p1l1+p2l2l1+l2Equal diameter means equal area In series Rnet=R1+R1pnet(l1+l2)A=p1+l1A+p2l2Apnet=p1l1+p2l2l1+l2

Two wires of equal diameters, of resistivities $p_{1}$ and $p_{2}$ and lengths $l_{1}$ and $l_{2}$ , respectively, are joined in series. The equivalent resistivity of the combination is

The correct option is A $\frac{p_{1} l_{1} + p_{2} l_{2}}{l_{1} + l_{2}}$
Equal diameter means equal area
In series
$R_{n e t} = R_{1} + R_{1} \frac{p_{n e t} (l_{1} + l_{2})}{A} = \frac{p_{1} + l_{1}}{A} + \frac{p_{2} l_{2}}{A} p_{n e t} = \frac{p_{1} l_{1} + p_{2} l_{2}}{l_{1} + l_{2}}$