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four copper wires of equal length are connected in series. their cross-sectional areas are 1.08 cm2, 1.94 cm2, 3.08 cm2, and 5.06 cm2. a potential difference of 120 v is applied across the combination. determine the voltage across the 1.94 cm2 wire.

Sagot :

bharathparasad577

The voltage applied across 1.94 cm² wire is22.1 V

We are given that

A₁ = 0.7 cm² = 1.08*10⁻⁴ m²

A₂= 2.5 cm² = 1.94*10⁻⁴ m²

A₃= 2.2 cm² = 3.08*10⁻⁴ m²

A₄= 3 cm² = 5.06*10⁻⁴ m²

Using  1cm²=10⁻⁴ m²

We know that using resistance formula we can calculate voltage

R = ρL/A

In series

R=R₁+R₂+R₃+₄

R=ρL/{A₁+ρL/{A₂}+ρL/{A₃}+ρL/{A₄}

Substitute the values

R = ρL/(2.62*10⁴)

V = 120v

I = V/R

I = 120 / {ρL(2.62*10⁴)}

Now the voltage across can be determined.

Voltage across the 1.94 square cm wire=IR=I*(ρL/A₂)

Voltage across the 1.94 square cm wire=ρL/ 1.94*10⁻⁴=22.1 V

Voltage across the 1.94 square cm wire=22.1 V

Therefore, The voltage applied across 1.94 cm² wire is22.1 V

To know more about voltage, refer: https://brainly.com/question/15293363

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