The figure shows a resistor of resistance R = 6.06 ohm connected to an ideal bat

Question

The figure shows a resistor of resistance R = 6.06 ohm connected to an ideal battery of emf E = 13.1 V by means of two copper wires. Each wire has length 21.2 cm and radius 2.50 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? Number Units Number Units Number Units Number Units Number Units

Explanation / Answer

The resistance fo the wires will be:

R = rho L/A

R = 1.68 x 10^-8 x 0.212/3.14 x 0.0025^2 = 0.0002 Ohm

for two wires in series

R' = 2R = 0.0004 Ohm

Now these are in series with 6.06

Req = 6.06 + 0.0004 = 6.0604 Ohm

The same current flows through the resistances connected in series

I = 13.1/6.0604 = 2.162 A

a)drop across R

V(r) = IR = 2.162 x 6.06 = 13.1 V

Hence, V(r) = 13.1 V

b)V(w) = I R = 2.162 x 0.0002 = 0.0004324

Hence, V(w) = 4.324 x 10^-4 V = 0.4324 mV

c)P(r) = I^2 R

P(r) = 2.162^2 x 6.06 = 28.326 W

Hence, P(r) = 28.326 W

d)P(w) = I^2 R

P(w) = 2.162^2 x 0.0002 = 0.0044 W

Hence, P(w) = 4.4 x 10^-3 W = 4.4 mW

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