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

L1xL2/(L1+L2)

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L1+L2

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L1-L2

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1/L1 + 1/L2

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Solution

The correct option is B
L1+L2

Resistance of both the resistance will be $R_{1}$ = $\frac{P L_{1}}{A}$ and $R_{2}$ = $\frac{P L_{2}}{A}$ .Since they are in series so $R_{n e t}$ = $\frac{P L_{n e t}}{A}$ = $R_{1}$ + $R_{2}$ = $\frac{p (L_{1} + L_{2})}{A}$ .So $L_{n e t}$ = $L_{1}$ + $L_{2}$ .

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In a circuit two resistances of the same material and cross section area but different length L1 & L2 are kept in series. Find the length of the equivalent resistance if it’s made of the same material and same cross section area.

The correct option is B L1+L2 Resistance of both the resistance will be R1 =PL1A and R2 =PL2A .Since they are in series so Rnet = PLnetA = R1 +R2 = p(L1+L2)A.So Lnet = L1 +L2.

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

The correct option is B
L1+L2

Resistance of both the resistance will be $R_{1}$ = $\frac{P L_{1}}{A}$ and $R_{2}$ = $\frac{P L_{2}}{A}$ .Since they are in series so $R_{n e t}$ = $\frac{P L_{n e t}}{A}$ = $R_{1}$ + $R_{2}$ = $\frac{p (L_{1} + L_{2})}{A}$ .So $L_{n e t}$ = $L_{1}$ + $L_{2}$ .