11th Physics | Quiz: Resistance and Ohm's Law

*(#8)* A silver wire is 4.5 m long and 0.45 mm in diameter. What is its resistance?

silver resistivity: 1.59 × 10⁻⁸ Ωm L: 4.5 m d: 0.45 mm → 45 × 10⁻³ m [45 × 10⁻³]/[2]=r=225 × 10⁻³ m A=πr² A=(3.14)(225 × 10⁻³)² A=1.59 × 10⁻⁷ m² R=[ρL]/[A] R=[(1.59 × 10⁻⁸)(4.5)]/[1.59 × 10⁻⁷] R=*0.45 Ω*

*(#10)* The tungsten filament of a light bulb has a resistance of 0.07 Ω. If the filament is 27 cm long, what is its diameter?

tungsten resistivity: 5.60 × 10⁻⁸ Ωm R: 0.07 Ω L: 27 cm → 27 × 10⁻² m R=[ρL]/[A] 0.07=[(5.60 × 10⁻⁸)(27 × 10⁻²)]/[A] A=2.16 × 10⁻⁷ m² [2.16 × 10⁻⁷]/[3.14]=6.88 × 10⁻⁸ √[6.88 × 10⁻⁸]=r=2.62 × 10⁻⁴ [2.62 × 10⁻⁴]2=d=*5.24 × 10⁻⁴ m*

*(#13)* A current of 0.76 A flows through a copper wire 0.44 mm in diameter when it is connected to a potential difference of 15 V. How long is the wire?

I: 0.76 A copper resistivity: 1.72 × 10⁻⁸ Ωm d: 0.44 mm → 44 × 10⁻3 m V: 15 V [44 × 10⁻³]/[2]=r=22 × 10⁻³ m V=IR → R=[V]/[I] R=[15]/[0.76] R=19.74 Ω A=πr² A=(3.14)(22 × 10⁻³)² A=15.20 × 10⁻⁴ m² R=[ρL]/[A] 19.74=[(1.72 × 10⁻⁸)(L)]/[15.20 × 10⁻⁴] 30.00 × 10⁻³= (1.72 × 10⁻⁸)(L) L=*174 × 10⁴ m*

Resistance formula

R: resistance (Ω) ρ: rho; resistivity (Ωm) L: length (m) A: area (m²)

*(#9)* When a potential difference of 18 V is applied to a given wire, it conducts 0.35 A of current. What is the resistance of the wire?

V: 18 V I: 0.35 A V=IR → R=[V]/[I] R=[18]/[0.35] R=*51.43 Ω*

Resistance and area are?

inversely proportional (when ↑R, ↓A)