A conducting wire has a 1.00 mm diameter, a 1.50 m length, and its resistivity i

Question

A conducting wire has a 1.00 mm diameter, a 1.50 m length, and its resistivity is 3.00 times 10^-7 Ohm middot m. This wire is connected between the terminals of a 300 V battery. The resistance of the conducting wire is A. 3.00 Ohm B. 1.50 Ohm C. 0.573 Ohm D. 0.143 Ohm Following Question 9, the electric power consumption of the conducting wire is A. 3.00 W B. 6.00 W C. 15.7 W D. 643 W Refer to the figure below for Question 11 to 13. Given R = 80.0 Ohm and V = 9.00 V the equivalent resistance of the circuit connected to the battery is A. 130 Ohm B. 96 Ohm C. 53 Ohm D. 50 Ohm The electric power provided by the battery is A. 1.01 W B. 0.623 W C. 0.844 W D. 1.53 W The potential difference between points A and B is A. 1.50 V B. 3.00 V C. 1.39 V D. 0.65 V

Explanation / Answer

here,

11)

R = 80 ohm

V = 9 V

(R+R) and R are in parallel, their equivalent resistance , R1 = 2 * R * R /( 2R + R) = 0.67 * R

(R1 + R) and R are in parallel, their equivalent resistance , R2 = ( R1 +R) * R /( R1 + R + R)

R2 = 1.67 * R * R /( 2.67)

R2 = 0.625 * R

the equivalemt resistance , Req = R2 + R = 1.625 * R = 1.625 * 80 = 130 ohm

12)

the electric power , P = V^2 /Req

P = 0.623 W

the electric power is B) 0.623 W

13)

the potential difference A and C, Vac = V - R * V/Req

Vac = 9 - 80 * 9 /130 V

Vac = 3.46 V

Vab = Vac - (V/Req - Vac /R) * R

Vab = 3.46 - ( 9/130 - 3.46/80) * 80

Vab = 1.39 V

the potential difference between
A and B is c) 1.39 V

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