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Faraday's second law states that the masses of different substances deposited or liberated by the same quantity of electricity are directly proportional to the chemical equivalents of the substances.

The capacitance of a capacitor, C, is inversely proportional to: Distance between plates: Capacitance decreases as the distance increases.

Capacitance in parallel = one at the top + one under = 2C The two in the middle are in series = \(\frac{C}{2}\) The equivalent capacitance of the capacitors in the circuit above = \(\frac{C}{2}\) + 2C = \(\frac{5}{2}\)C

The simple form of the lead-acid accumulator typically has a negative pole made of lead (Pb). In this type of battery, the negative electrode is composed of spongy lead, while the positive electrode is made of lead dioxide (PbO\(_2\))

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

Efficiency = \(\frac{\text{ Output energy}}{\text{input energy}}\) x 100 Efficiency = 90% = 0.9, Output energy = 2700J Input energy = \(\frac{2700}{0.9}\) = 3000J

mass of metal deposited during electrolysis = Zit Given: mass = 0.02kg, i, t = 120s and Z = 1.3 x 10\(^{-7}\)kgC\(^{-1}\)] i = \(\frac{\text{m}}{\text{Zt}}\) i = \(\frac{0.02}{1.3 \times 10^{-7} \times 120}\) = 1282.05 ≈ 1.3 x 10\(^3\)A

Dry Leclanché: This is a common non-rechargeable cell.

To find the value of R required to make the galvanometer measure up to 40 V, we can use the following formula for a series circuit involving a galvanometer V = I ( R + r ) R = \(\frac{\text{V}}{\text{I}}\) - r where R = resistance in series = ?, V = total voltage = 40V, I = current through galvanometer = 20Am = 0.02A and r = internal resistance = 10\(\Omega\) R = \(\frac{40}{0.02}\) - 10 R = 2000 - 10 = 1990\(\Omega\)

A potentiometer is preferred over a voltmeter for voltage measurements because it does not draw significant current from the circuit being measured.

Power = \(\frac{\text{Energy consumed}}{\text{time}}\) 150W → 0.150KW, time = 336hrs. Energy consumed in KWh = 0.150 x 336 = 50.40KWh

The balance length of a potentiometer is directly proportional to the e.m.f Thus, \(\frac{L_1}{L_2} = \frac{E_1}{E_2}\) \(L_2 = L_1\times \frac{E_2}{E_1}\) \(L_2 = 75 \times \frac{2.295}{3.06}\) = 75 x 0.750 = 56.25cm

Given: I\(_g\) = 50mA = 0.05A, I to be measured = 2A, r = 2\(\Omega\), I\(_s\) = I - I\(_g\) = 2 - 0.05 = 1.95A Shunt(R) = \(\frac{I_g}{I_s}\) x r R = \(\frac{0.05}{1.95}\) x 10 = 0.2564\(\Omega\)

Total power consumption = 80 + 100 + 120 = 300W P in KW = 0.3KW Energy consumed = 0.3 x 6 = 1.8KWh Monthly energy consumption = 1.8 x 31 = 55.8KWh Cost of Monthly energy usage = 55.8 x 24.08 = #1343.664k ( since 1kwh = #24.08k)

At resonance, the circuit's impedance is at its lowest (equal to the resistance), which results in the highest current flow. This condition is achieved when the inductive reactance and capacitive reactance are equal, causing them to cancel each other out.