Figure 27-30 shows a resistor of resistance R = 6.01 ? connected to an ideal bat

Question

Figure 27-30 shows a resistor of resistance R = 6.01 ? connected to an ideal battery of emf = 13.7 V by means of two copper wires. Each wire has length 22.4 cm and radius 4.00 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 (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?

Figure 27-30 shows a resistor of resistance R = 6.01 ? connected to an ideal battery of emf = 13.7 V by means of two copper wires. Each wire has length 22.4 cm and radius 4.00 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 (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?

Explanation / Answer

Solution

voltage/Potential difference across resistor=13.7V(ideal case)

B).13.7V(in case of ideal)

c).

thermal energy lost

V=I2R

I=v/R

13.7/6.01=2.28A

So, I2R=31.2296W

D0.

resistance Rw=L/A

A=r2

3.1415*4*4=50.26

A=50.26

Rw=L/A

Rw=1.7 x 10-8 x22.4/50.26

Rw=0.7576x10-8

I2R across Rw=

Iw=V/Rw

13.7/0.7575x 10^-8

Iw=0.18A

0.0324x0.7576=0.24546x10-14W

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