ire rry s is lec- 4. Three wires are made of copper having circular cross sectio

Question

ire rry s is lec- 4. Three wires are made of copper having circular cross sections. Wire I has a length I. and radius Wire 2 has a length I and radius 2r Wire 3 has a length 21 and radius 3r. Which wire has the smallest resistance? (a) wire 1 (b) wire 2 (c) wire 3 (d) All three wires have the same resistance. (e) Not enough information is given to answer the question nd nce 5. Which of the following combinations of units has units of energy? (a) C/s (b) kWh (c) kW (d) A h (e) W/s he is ed 9. A color television set draws about 2.5 A when con- nected to a 120 V source. What is the cost (with elec- trical energy at 8 cents/kWh) of running the set for 8.0 hours? (a) 20 cents (b) 4.0 cents (e) 19 cents (d) 40 cents (e) 62 cents 10. A potential difference of 1.0 V is maintained across a 10.0- resistor for a period of 20 s. What total charge passes through the wire in this time interval? (a) 200 C (b) 20 C (c) 2 C (d) 0.005 C (e) 0.05c 11. Three resistors, A, B, and C, are connected in parallel and attached to a battery, with the resistance of A being the smallest and the resistance of C the greatest. Which resistor carries the highest current? (a) A (b) B (c) C (d) All wires carry the same current. (e) More informa- tion is needed to answer the question. 12. Three resistors, A, B, and C, are connected in series in a closed loop with a battery, with the resistance of A lbeing the smallest and the resistance of C the greatest. Across which resistor is the voltage drop the greatest? (a) A (b) B (c) C (d) The voltage drops are the same for each. (e) More information is needed to answer the question. 13. The power delivered to a resistor is 4.0 W when a cer- tain voltage is applied across it. How much power is delivered to the resistor when the voltage is doubled? (a) 2.0 W (b) 4.0 W (c) 8.0 W (d) 16 W (e) 32 W

Explanation / Answer

At one time I will solve first 4 questions according to Q&A guidelines.

4) Resistance = R = p*L/A

For wire 1 ========> R1 = p*L/(pi*r^2)

For wire 2 ========> R2 = p*L/(4pi*r^2) = R1/4

For wire 3 ========> R3 = p*2L/(9pi*r^2) = R1/4.5

So, wire 3 has the smallest resistance.

5) b) kWh

9) I = 2.5 A

    V = 120 V

    P = V*I = 120*2.5 = 300 W

   Unit = P*t = 300W*8h = 2.4 kWh

Cost = 2.4*8 = 19.2 cents (Option C)

10) V = 1.0 V

       R = 10 ohm

      I = V/R = 1/10 = 0.1 A

     t = 20 s

     Charge = Q = I*t = 0.1*20 = 2 C (Option C)

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