Question

12. What work is done when 5.0 C is raised in potential by 1.5 V.......

13. What potential difference is applied to metal plates 0.500 m apart if the electric field between them is 2.50 x 103N/C.

14. In ohmic material one of this option is not correct (a)Vul (b) Slope is equal to resistance (3) Straight line runs from the ori in (d) None of the above

15. A voltmeter measures the potential difference between two large parallel plates to be 60.0 V. the plate are 3.0 cm apart.

What is the magnitude of the electric field between them.... 16. Potential drop across the wire is directly proportional to 17. With three resistance in parallel R₁=70, R2 = 90 and

R3=120, Calculate the 11 and 12 when I supplied to the circuit is.....

Answer 1

[AI-generated answer — Bot GPT 3.5]

12. The work done would be equal to QΔV, so W = (5.0 C)(1.5 V) = 7.5 J.

13. The potential difference applied to the metal plates can be calculated using the formula ΔV = Ed, where E is the electric field strength and d is the distance between the plates. So, ΔV = (2.50 x 10^3 N/C)(0.500 m) = 1250 V.

14. Option (d) None of the above is correct for an ohmic material.

15. The magnitude of the electric field between the plates can be found using the formula E = V/d, where V is the potential difference and d is the distance between the plates. Therefore, E = 60.0 V / 0.030 m = 2000 V/m or 2.0 x 10^3 N/C.

16. The potential drop across a wire is directly proportional to the current flowing through it, according to Ohm's Law (V = IR).

17. The equivalent resistance of resistors in parallel can be calculated using the formula 1/Req = 1/R1 + 1/R2 + 1/R3. Therefore, 1/Req = 1/70 + 1/90 + 1/120 = (36/630) + (35/630) + (28/630) = 99/630. Therefore, Req = 630 / 99 = 6.36 ohms. The total current supplied to the circuit is given as I, so the potential drop across the resistors R1 and R2 can be calculated using Ohm's Law (V = IR). V1 = (70)(I) and V2 = (90)(I).