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C. Ammonium chloride. It is the electrolyte in the wet Leclanché cell (aqueous solution).Others: Nickel hydroxide: In Ni-Cd/Ni-MH batteries. Lead dioxide: Cathode in lead-acid batteries. Carbon: Conductivity enhancer, not electrolyte.

Parallel connection does not add the voltages; it only increases the current capacity. Thus, total e.m.f. E = E\(_1\) = E\(_2\) = E\(_3\).

E = \(\frac{1}{2}\)QV; But, Q = CV ∴ E = \(\frac{1}{2}\)CV\(^2\); 40 = \(\frac{1}{2}\) \times 5 \times 10\(^{-6}\) \times V\(^2\) V\(^2\) = \(\frac{(2 \times 40)}{5 \times 10^{-6}}\) V\(^2\) = 16 \times 10\(^6\) ∴ V = \(\sqrt{(16 \times 10^6)}\) = 4000 V

The quantity of charge flowing through a conductor is given by the formula Q = It. Substituting the given values, we get Q = 10 \times 10 = 100 C.

\(\frac{1}{R_{XY}}\) = \(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}\) = \(\frac{5}{6}\) \(\therefore\) R\(_{XY}\) = \(\frac{5}{6}\)\(\Omega\) \(\therefore\) R\(_T\) = \(\frac{5}{6}\) + 0.8 = \(\frac{5}{6}\) + \(\frac{4}{5}\) = \(\frac{10}{5}\) = 2\(\Omega\) But, V = IR\(_T\) where V = 5v and R\(_T\) = 2\(\Omega\) I = \(\frac{\text{V}}{R_T}\) = \(\frac{5}{2}\)A \(\therefore\) V\(_{across XY}\) = I R\(_{XY}\) = \(\frac{5}{2}\) x \(\frac{6}{5}\) = 3.0V.

V = IR Voltage across XY = 3V V = 3I = 3V I = \(\frac{3}{3}\) = 1.0 A

The resistance \(R\) of a wire is related to its resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) by the formula: \(R = \frac{\rho L}{A}\) The conductivity \(\sigma\) is the reciprocal of the resistivity: \(\sigma = \frac{1}{\rho}\) \(\rho = \frac{R A}{L}\) \(\sigma = \frac{L}{R A}\) Substitute the given values (\(L = 5\) m, \(R = 10\) \Omega, \(A = 4.0 \times 10^{-3}\) m²): \(\sigma = \frac{5}{10 \times 4.0 \times 10^{-3}} = \frac{5}{0.04} = 125\) The conductivity of the wire is 125 \Omega\(^{-1}\)m\(^{-1}\) = 1.25 x 10\(^2\) \Omega\(^{-1}\)m\(^{-1}\)

3a. The e.m.f of the accumulator = 1.5V Tables of values R(\(\Omega\)) I(A) I\(^{-1}\)((A\(^{-1}\)) 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 S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{4 - 2}{4 - 1}\) = \(\frac{2}{3}\) = 0.67 The intercept on the vertical axis = 1.32\(\Omega\) Precautions: (i) I ensured all electrical connections were securely tightened to prevent loose contacts, which could lead to fluctuating or inaccurate readings. (ii) I avoided parallax error by positioning my eyes directly in line with the scale when reading the ammeter or voltmeter for precise measurements. (iii) I checked and corrected for any zero error on the ammeter and voltmeter before taking readings to improve accuracy. (iv) I removed the key from the circuit when not taking measurements to avoid unnecessary heating and potential damage to components. (v) I took multiple readings at each data point and averaged the values to minimize random errors and ensure more reliable results. (vi) I ensured the apparatus was properly calibrated and regularly checked for consistency throughout the experiment. bi. 1. Rechargeability: A lead-acid accumulator is rechargeable, making it more suitable for long-term use, whereas a Leclanché cell is primarily a primary (non-rechargeable) cell. 2. Higher Current Supply: Lead-acid accumulators can deliver higher currents, making them ideal for applications requiring significant power, like in vehicles, while Leclanché cells are suitable for low-current devices. 3. Longer Lifespan: Due to its rechargeable nature, the lead-acid accumulator has a longer operational life compared to a Leclanché cell, which has a limited lifespan and must be replaced after depletion. 4. Stable Voltage: Lead-acid accumulators provide a relatively stable output voltage during discharge, while the voltage of a Leclanché cell drops significantly over time. bii. Since 3 \Omega and 4 \Omega are connected in parallel, \(\frac{1}{\text{R}}\) = \(\frac{1}{3}\) + \(\frac{1}{4}\) \(\frac{1}{\text{R}}\) = \(\frac{4 + 3}{12}\) \(\frac{1}{\text{R}}\) = \(\frac{7}{12}\) R = \(\frac{12}{7}\) = 1.71 \Omega Now, 1.71 \Omega is connected in series with 4 \Omega. ∴ The effective resistance in the circuit = 1.71 \Omega + 4 \Omega = 5.71 \Omega

Answer: B. 2.27A (I = \(\frac{\text{P}}{\text{V}}\) = \(\frac{500}{220}\) ≈ 2.27 A.)

C. 50.40Kwh Convert power to kW (150 W = 0.15 kW), then multiply by time in hours: energy = 0.15 kW \times 336 h = 50.4 kWh.

The circuit is a simple series circuit with a 16 V battery (or voltage source) directly connected across a 2 \Omega resistor. The power (\(P\)) dissipated (developed) in the resistor is given by: \(P = \frac{V^2}{R}\) where \(V = 16 \, \text{V}\) and \(R = 2 \, \Omega\). Calculating the power: \(P = \frac{16^2}{2} = \frac{256}{2} = 128 \, \text{W}\)

For an A.C. current, \(I_{\text{r.m.s}} = \frac{I_0}{\sqrt{2}}\) where (I_0) is the maximum (peak) current. Given (I_0 = 7,\text{A}), \(I_{\text{r.m.s}} = \frac{7}{\sqrt{2}} \approx 4.95,\text{A}\) Answer: \(4.95 \text{A}\ \ (\approx 5\ \text{A})\)

Given: Inductance: \( L = 0.5 \, \text{H} \), Resistance: \( R = 45 \, \Omega \), Frequency: \( f = 50 \, \text{Hz} \) inductive reactance \( X_L \): \(X_L = 2 \pi f L = 2 \times \frac{22}{7} \times 50 \times 0.5 = \frac{1100}{7} \approx 157.14 \, \Omega\) The impedance \( Z \): \(Z = \sqrt{R^2 + X_L^2} = \sqrt{(45)^2 + (157.14)^2} = \sqrt{2025 + 24642.82} \approx \sqrt{26667.82} \approx 163.5 \, \Omega\).

The period of a wave is the reciprocal of its frequency. Therefore, period = \(\frac{1}{\text{frequency}}\) Substituting the given frequency, we get period = \(\frac{1}{ 50}\) Hz = 0.02 s.

V = I X\(_L\) X\(_L\) = \(\frac{\text{V }}{\text{I}}\) = \(\frac{240}{2}\) = 120 \Omega But X\(_L\) = 2\(\pi\)fL 120 = 2 x 2\(\pi\) x \(\frac{100}{\pi}\) x L 120 = 2 x 100 x L = 200L L = \(\frac{120}{200}\) = 0.6H