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M = Zit Z = \(\frac{M}{It}\) = \(\frac{0.48 \times 10^{-3}}{4 \times 10^2 \times 3600}\) = 3.3 x 10\(^{-10}\)kgc\(^{-1}\)

The e.m.f of the accumulator = 1.5v Table of value/observations R(\(\Omega\)) 1\(_{amp}\) 1\(^{-1}_{amp}\) 0 0.78 1.28 1 0.50 2.00 2 0.38 2.63 3 0.30 3.33 4 0.25 4.00 5 0.22 4.55 Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{4-1}{4-2} = \frac{3}{2}\) The intercept on the vertical axis is =-1.85\(\Omega\) Precautions: I tightened connections. I avoided errors due to parallax in reading ammeter or voltmeter. I noted zero error of ammeter/ voltmeter. I removed key when readings were not taken. I took repeated readings from the table, i.e. averaging the reading. (b)i. - The lead-acid accumulator can be recharged while the Leclanche cell cannot. - The lead-acid accumulator has the capacity to supply large currents for a long time while the leclanche cell does not. ii. The two resistors 3\(\Omega\) and 4\(\Omega\) are connected in parallel. \(\therefore\) The equivalent resistance (\(\Omega\) = \(\frac{1}{R} = \frac{1}{3} = \frac{1}{4} = \frac{4+3}{12}\) = \(\frac{7}{12}\) = 1.71 The effective resistance in the circuit = 1.71 + 4 = 5.71\(\Omega\)

Since all the cells are connected in parallel, the effective e.m.f. is just that of one single cell. Which means e.m.f. = 1.5V However, the effective internal resistance adds as parallel arrangement given as 1/r = 1/r1 + 1/r2 + 1/r3 + 1/r4 = 1/4 + 1/4 + 1/4 + 1/4 = 1\Omega ∴ r = 1\Omega Answer = 1.5V, 1\Omega

The maximum current drawn by the cooker is given by; I = \(\frac{P}{V}\) ∴ I = \(\frac{\text{3000}}{200}\) I = 15A The fuse rating = I \(\times\) 125% (the fuse is rated at 125% of the normal operating current) Hence, the fuse rating = 15A \(\times\) (\(\frac{125}{100}\)) = 18.75amps Therefore, the best fuse is 20A.

Shunt resistance is usually given as R = igrg/(I - ig) Where R = 0.05Ω rg = 5Ω I = 2M ∴ ig = IR/(rg + R) = (2 x 0.05)/(5 + 0.05) = 0.10/5.05 = 0.0198A = 19.8mA = 20mA

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

A rectifier of an alternating current is a device which allows the flow of current only in one direction

The effective V = √ VR2 + VL2  = √ 62 + 82  = √ 36 + 64  = √ 100  10V

When a magnet is moved near a ferromagnetic material like soft iron, it induces magnetism in the material. The end closer to the magnet becomes the opposite pole. In this scenario, the south pole of the magnet X approaches point P. Point P becomes a north pole to repel the south pole, while the opposite end (Q) becomes a south pole.

From Ohms Law, E = I(R +X) => E = I1(R1 + X) again, E = I2(R2 + X) => I1(R1 + X) = I2(R2 + X) ∴ 5(8 + X) = 7(5 + X) 40 + 5X = 35 + 7X 7X - 5X = 40 - 35 2X = 5 X = 5/2 = 2.5Ω

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 A.

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

Mass deposited = 21t I = \(\frac{m}{2t} = \frac{0.66}{33 \times 10^{-5} \times 30 \times 60}\) = 1.11A error = 1.11 - 1 = 0.11A

(a) This is the process by which a solution or a molten substance is broken down into the component parts by the passage of an electric current through it. (b) \(\frac{Electrolytes}{\text{salt solution}}\) = \(\frac{\text{non - electrolytes}}{\text{Sugar solution}}\) grape fruit juice paraffin