First slide

Resistivity of various materials

Question

Two rods are joined end to end, as shown. Both have a crosssectional area of $0.01 c m^{2}$ . Each is 1 meter long. One rod is of copper with a resistivity of $1.7 \times 10^{- 6}$ ohm-centimeter, the other is of iron with a resistivity of $10^{- 5}$ ohm-centimeter. How much voltage is required to produce a current of 1 ampere in the rods?

Moderate

A

$0.00145 V$
B

$0.0145 V$
C

$1.7 \times 10^{- 6} V$
D

$0.117 V$

Solution

$H e r e, L e n g t h o f e a c h r o d, l = 1 m Area of cross-section of each rod, A = 0.01 {cm}^{2} = 0.01 \times 10^{- 4} m^{2} Resistivity of copper rod, ρ_{cu} = 1.7 \times 10^{- 6} Ω cm = 1.7 \times 10^{- 6} \times 10^{- 2} Ω m = 1.7 \times 10^{- 8} Ω m Resistivity of iron rod, ρ_{Fe} = 10^{- 5} Ωcm = 10^{- 5} \times 10^{- 2} Ωm = 10^{- 1} Ωm ∴ Resistance of copper rod R_{Cu} = ρ_{Cu} \frac{l}{A} and resistance of iron rod, R_{Fe} = ρ_{Fe} \frac{l}{A} As copper and iron rods are connected in series, therefore equivalent resistance is R = R_{Cu} + R_{Fe} = ρ_{Cu} \frac{l}{A} + ρ_{Fe} \frac{l}{A} = (ρ_{Cu} + ρ_{Fe}) \frac{l}{A}$
$Voltage required to produce 1 A current in the rods is V = I R = (1) (R_{Cu} + R_{Fe}) = (ρ_{Cu} + ρ_{Fe}) (\frac{l}{A}) = (1.7 \times 10^{- 8} + 10^{- 7}) (\frac{1}{0.01 \times 10^{- 4}}) V = 10^{- 7} (0.17 + 1) (10^{6}) V = 1.17 \times 10^{- 1} V = 0.117 V$

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