Question Library

R = \(\frac{4eL}{\pid^2}\) L = \(\frac{R\pid^2}{4e}\) = \(\frac{21 \times 3.14(4 \times 10^{-3})^2}{4 \times 3.0 x 10^{-8}}\) = 8.8 x 103m

E = \(\frac{v}{d} = \frac{10^3}{0.1}\) = 10000 = 1.0 x 104Vm-1

No expanded working has been attached for this question yet. The verified answer is Option B.

(a) Doping is the introduction of impurity atoms into a semicondoctor (b) Doping increases the conductivity of a semicondoctor by increasing the number of chargers or electron or holes.

Magnetic relay device :is the device that controls a circuit (i.e one circuit )if a current in the second circuit is large or dangerous

Electron: A sub–atomic particle haring a negative charge and orbiting the nucleus (note: the symbol 0–1e represent the particle of an electrons.)

A good electric cell should proses the following 1. A long life of an electric cell, 2. A minimum of polarization effect 3. A cheap and safe electrolytes Except D cost and dangerous of the electric cell

Vr.m.s = V0 ÷ V2, where Vr.m.s rep. to as voltage root –square mean, v = peak voltage Vr.m.s = 200volt, v = ? 200 = V0 ÷ V2 V0 = 200v2 = 200 × 1.414 = 282.8 volts V0 = b &approx.; 283 volts

E - V1 + V2 24V = 14V + VQP VQP = 4 + 14V VQP =10V

No expanded working has been attached for this question yet. The verified answer is Option A.

R = \(\frac{PL}{A}\) C = \(\frac{RA}{P}\) C = \(\frac{0.60 \times 10^{-6} \times 3}{7.2 \times 10^{-7}}\) C = 2.50m

E - V1 + V2 24V = 14V + VQP VQP = 24 + 14V VQP = 10V

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option C.

(a) Doping in a semi conductor is the addition of impurity to a pure semi condutor to increase its electrical conductivity through increase in the number of holes and electrons or charge carriers.