Your answer is partially correct. Try again. The figure shows a resistor of resi

Question

Your answer is partially correct. Try again. The figure shows a resistor of resistance R = 6.25 connected to an ideal battery of emf E = 12.5 V by means of two copper wires. Each wire has length 22.3 cm and radius 3.10 mm. In dealing with such circuits in this chapter, we generally neglect the potential differences along the wires and the transfer of energy to thermal energy in them. Check the validity of this neglect for the circuit of the figure below. What is the potential difference across (a) the resistor and (b) each of the two sections of wire? At what rate is energy lost to thermal energy in (c) the resistor and (d) each section of wire? Wire 8 Wire 2 UnitsV (a) Number25 (b) Number TT 24.8 (c) Number T 50 (d) Number49.6 Units mv Units w Units mW

Explanation / Answer

resistance = resistivity * length / area

resistivity of copper = 1.7 × 10^-8 ohm m

resistance of 22.3 cm wire = 1.7 * 10^-8 * 0.223 / (pi * (3.10 * 10^-3)^2)

resistance of wire = 1.256 x 10-4 ohm

wires and resistances are in series so,

net resistance = 2 * resistance of wires + resistance of resistor

net resistance = 2 * 1.256 x 10-4 ohm + 6.25

net resistance = 6.25025 ohm

by ohm's law

V = IR

12.5 = I * 6.25025

I = 2 A

potential = resistance * current

potential across resistance = 6.25 * 2

potential across resistance = 12.5 V

potential across each of the two wire = 1.256 x 10-4 ohm * 2

potential across each of the two wire = 0.25 mV

thermal energy lost = current^2 * resistance

thermal energy lost by resistor = 2^2 * 6.25

thermal energy lost by resistor = 25 W

thermal energy lost by each wire = 2^2 * 1.256 x 10-4 ohm

thermal energy lost by each wire = 0.50 mW

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.