Respuesta :
Answer:
Therefore the voltage across the 1.9 cm² wire is 25.15 v.
Explanation:
Given that,the four copper wires is connected in series and their length of the copper wire are same.
The cross section area of the wires are 0.6 cm², 1.9 cm², 4.1 cm² and 5 cm².
The resistance of a wire depends the the length of wire, the cross section area of the wire and the resistivity of the wire.
[tex]R = \frac{\rho.l}A[/tex]
R = resistance of a wire
[tex]\rho[/tex]= resistivity of the wire
l = length of the wire
A= cross section area of the area.
Let the length of the wires be l and resistivity be [tex]\rho[/tex].
If the resistance connects in series, then the total resistance
R=R₁+R₂+R₃+....
Here the length of the wires and the resistivity of the wires are same for all wire, but the cross section areas are different for each of them
[tex]R_1= \frac{\rho l }{0.6}\times 10000 \ ohm[/tex] , [tex]R_2= \frac{\rho l }{1.9}\times 10000 \ ohm[/tex], [tex]R_3= \frac{\rho l }{4.1}\times 10000 \ ohm[/tex] and [tex]R_4= \frac{\rho l }{5}\times 10000 \ ohm[/tex]
Therefore the total resistance
[tex]R=\rho l[\frac 1{0.6}+\frac 1{1.9}+\frac 1{4.1}+\frac15] \times 10000[/tex] [ 10000 multiply convert cm² to m²]
[tex]=\rho l[\frac{10}6+\frac{10}{19}+\frac{10}{41}+\frac15] \times 10000[/tex]
[tex]=\rho l [\frac{38950+12300+5700+4674}{23370}]\times 10000[/tex]
[tex]=\rho l\frac{61,624}{23,370}\times 10000[/tex] ohm
The voltage of 126 v is applied to the arrangement.
Ohm's Law:
[tex]I=\frac VR[/tex]
I= current
R= resistance
V= voltage
Therefore the current passes through the circuit
[tex]I=\frac{126}{\rho l\frac{61,624}{23,370}\times 10000}[/tex] A
The voltage across the 1.9 cm² is
[tex]V=I\times R_2[/tex]
[tex]=(\frac{126}{\rho l\frac{61,624}{23,370}\times 10000})\times (\frac{\rho l \times 10000}{1.9})[/tex]
[tex]=\frac{126\times 23,370}{61,624\times 1.9}[/tex]
≈25.15 V
Therefore the voltage across the 1.9 cm² wire is 25.15 v.
The voltage across the 1.9 cm² wire is 25.25 V when a voltage of 126 V is applied to the series arrangement.
Calculating the voltage drop:
The relation between resistivity and resistance is expressed as follows:
R = ρl/A
where R is resistance
ρ is resistivity
l is the length of the conductor
A is the cross-sectional area
Given that all wires have the same length, say L.
The resistivity of copper is will ve the same for all, say ρ.
So, the resistances of the given wires of the cross-sectional area of 0.6cm², 1.9cm², 4.1cm² and 5 cm will be:
R₁ = ρL/0.6 , R₂ = ρL/1.9 , R₃ = ρL/4.1 , and R₄ = ρL/5
since they are connected in series with a voltage of 126 V, the voltage across the 1.9 cm² wire of R₂ will be:
V₂ = R₂V/(R₁+R₂+R₃+R₄)
V₂ = (ρL/1.9) × 126 / (2.637ρL)
V₂ = 25.15V
Learn more about resistivity:
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