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Alchen [17]
3 years ago
7

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

m2 . If a voltage of 140 V is applied to the arrangement, determine the voltage across the 1.2 cm2 wire. Answer in units of V.
Physics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

63.8 V

Explanation:

We are given that

A_1=1.6 cm^2=1.6\times 10^{-4} m^2

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

A_2=1.2 cm^2=1.2\times 10^{-4} m^2

A_3=4.4 cm^2=4.4\times 10^{-4} m^2

A_4=7 cm^2=7\times 10^{-4} m^2

Potential difference,V=140 V

We know that

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

According to question

l_1=l_2=l_3=l_4=l

In series

R=R_1+R_2+R_3+R_4

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

R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})

R=\rho l(18284.6)

I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}

Potential across 1.2 square cm=V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V

Hence, the voltage across the 1.2 square cm wire=63.8 V

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