Question

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 cm2 . 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.

Answer 1

63.8 V

Step-by-step explanation:

We are given that

`A_1=1.6 cm^2=1.6* 10^(-4) m^2`

`1 cm^2=10^(-4) m^2`

`A_2=1.2 cm^2=1.2* 10^(-4) m^2`

`A_3=4.4 cm^2=4.4* 10^(-4) m^2`

`A_4=7 cm^2=7* 10^(-4) m^2`

Potential difference,V=140 V

We know that

`R=(\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((1)/(A_1)+(1)/(A_2)+(1)/(A_3)+(1)/(A_4))`

`R=\rho l((1)/(1.6* 10^(-4))+(1)/(1.2* 10^(-4))+(1)/(4.4* 10^(-4))+(1)/(7* 10^(-4)))`

`R=\rho l(18284.6)`

`I=(V)/(R)=(140)/(\rho l* 18284.6)`

Potential across 1.2 square cm=
`V_1=IR_1=(140)/(\rho l* 18284.6)* \rho l((1)/(1.2* 10^(-4))=63.8 V`

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