The figure shows a resistor of resistance R = 6.21 ohm connected to an ideal bat

Question

The figure shows a resistor of resistance R = 6.21 ohm connected to an ideal battery of emf = 14.7 V by means of two copper wires. Each wire has length 21.8 cm and radius 2.00 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 the resistor and each of the two sections of wire? At what rate is energy lost to thermal energy in the resistor and each section of wire?

Explanation / Answer

The resistivity (rho) of the copper at room temperature is 1.678 x 10^-8 ohm-m.

R = resistivity * length / area ;

the length of each wire L is 0.218 m

A = PI * r^2 = 3.14 x 2^2 mm^2 = 12.56 mm^2 = 1.256 x 10^-5 m^2

Now:

R = (1.678 x 10^-8 x 0.218) / (1.256 x 10^-5)

R = 0.0002912 ohm

Total resistance of the circuit Rt is:

Rt = 2 * 0.00029 ohm + 6.21 ohm

Rt = 6.21058 ohm

Then the current I on the circuit is:

I = V / Rt = 14.7 V / 6.21058 ohm = 2.36692 A

(a)

V = IR

V(resistor) = 2.36692 A x 6.21 ohm = 14.6986 V

(b)

V(wire) = 2.36692 A x 0.0002912 ohm = 0.000689 V = 0.689 mV

(c)

W = V(resistor) * I = 14.6986 V x 2.36692 A = 34.79 watts

(d)

W = V(wire) * I = 0.00163 watts = 1.63 mW

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