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mezya [45]
3 years ago
15

The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

which the current takes different paths through a piece of material. Each of the rectangular pieces is made from a material whose resistivity is rho = 1.50 x 10-2 Ω*m, and the unit of length in the drawing is L0 = 7 cm. Each piece of material is connected to a 3.00-V battery. Find (a) the resistance and (b) the current in each case.
Physics
2 answers:
alex41 [277]3 years ago
5 0

Answer:

Resistance=0.5 ohms

Current=10 A

maria [59]3 years ago
4 0

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

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tresset_1 [31]

The definition of the celestial bodies allows us to find that the correct answer for a body that is captured and is in planetary orbit is:

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