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Veseljchak [2.6K]
3 years ago
14

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

m2 . If a voltage of 145 V is applied to the arrangement, determine the voltage across the 2.5 cm2 wire.
Physics
1 answer:
Travka [436]3 years ago
6 0

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

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