The figure shows a resistor of resistance R = 6.10 Ohm connected to an ideal bat

Question

The figure shows a resistor of resistance R = 6.10 Ohm connected to an ideal battery of emf epsilon = 15.4 V by means of two copper wires. Each wire has length 20.5 cm and radius 1.50 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? (a) Number Units (b) Number Units (c) Number Units (d) Number Units

Explanation / Answer

the resistances of two wires are,

   R1 = R2 = rho*L/A = rho*L/pi*r2 = 1.69x10-8 * 20.5x10-2 / pi*[1.5x10-3]2 = 4.90e-4 Ohm

equivalent resistance:

R = R1+R2+6.10 = 4.90e-4 + 4.90e-4 + 6.10 = 6.10098 Ohm

current in the circuit:

   I = V/R = 15.4 / 6.10098 = 2.5241845 A

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potential difference across resistor:

   VR = IR = 15.3975 V

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potential difference across each resistor:

   VR = IR = 2.5241845 *4.90e-4 =  1.2368e-3 V

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rate of loss of energy:

   P = V2/R = [15.3975]2 / 6.10 = 38.866 W

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rate of loss of energy:

   P = V2/R = [1.2368e-3]2 / 4.90e-4 = 3.12e-3 W = 3.12 mW

  

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