Physics Questions and Answers

Electric Power = IV = I²R = ^V2/_R thus for a device rated 1000 W, 230V, from the relation P = ^V2/_R, 1000 = ⁽²³⁰⁾²/_R
=> R = ⁽²³⁰⁾²/₁₀₀₀
= 52900/1000 = 52.9Ω

The iris controls the amount of light by limiting the volume of light admitted. The lens then focuses the light onto the retina, forming an inverted image. The lens changes its shape to adjust the focus on both distant and near objects, a process called accommodation.

This adjustment is done by two sets of muscles in the iris

1 KW = 1000 watts, power P = IV, thus for a mains of 250 V, the maximum permissible Current I = ¹/_V = 1000/250 = 4 A
Thus the fuse rating in the plug should be 4 A

The energy stored in a given capacitor is given by
E = ¹/₂ CV² = ¹/₂ QV = ¹/₂Q²/C
Where C = the capacitance of the capacitor
Q = charge carried by the capacitor
V = p.d across the plates of the capacitor thus from the data given, C = 10μF
= 10 x 10^-6F
Q = 100μc = 100 x 10^-6C
∴ Energy stored = ¹/₂ Q²/C

Let the combined parallel resistance of 3 Ω and x = Q Ω
∴ Total resistance in the circuit = 2 + Q + 1.5 = 3.5 + Q Ω
Total current in the circuit = 2A
P.D a cross the circuit 12 volts
∴ From Ohm’s law, V = IR
12 = 2(3.5 + Q)
= 7.0 + 2Q
∴ 2Q = 12 – 7
= 5
∴ Q = ⁵/₂ = 2.5 Ω
Thus from parallel formula: I/Q = 1/3 + 1/x
¹/_2.5 = ¹/₃ + ¹/_x
∴ ¹/_x = ¹/_2.5 - ¹/₃ = ^3-2.5/_7.5 = ^0.5/_7.5
X = ^7.5/_0.5 = ⁷⁵/₅ = 15 Ω
∴ X = 15 Ω