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Efficiency = Output/input x 100/1 Efficiency = 3/(3+1) x 100/1 = 3/4 x 100/1 = 75%

Since capacitance C = Q/V, => Q = CV ∴ Qp = 5 x 10\(^{-6}\) x 20 = 100 x 10\(^{-6}\) or 100\muC Qp = 10 x 10\(^{-6}\) x 20 = 200 x 10\(^{-6}\) or 200\muC Correct option 100\muC and 200\muC

Effective resistance in the circuit = 3 + 2 + 1 = 6\Omega Current through the I = \(\frac{E}{(R+r)}\) ∴ I = \(\frac{2}{6}\) => V1 = I x R1 = \(\frac{2}{6}\times\frac{3}{1}\) = 1V = V2 = I x R2 = \(\frac{2}{6}\times\frac{2}{1}\) = \(\frac{2}{3}\)V ∴ we have 1V and \(\frac{2}{3}\)V

In the provided circuit diagram, the correct position for a fuse is between the "Line" connection and the component labeled "P". The purpose of a fuse is to protect the circuit and the connected device from excessive current flow that could potentially cause damage. By placing the fuse at position P, it will be in series with the "live" or "hot" wire, allowing it to interrupt the circuit if the current exceeds the fuse's rated capacity.

Impedance Z = \(\sqrt{R^2 + X^2}\) = \(\sqrt{(30)^2 + (40)^2}\) ∴ Z = \(\sqrt{2500}\) = 50\(\Omega\) But current I = \(\frac{V}{Z}\) = \(\frac{200}{50}\) = 4.00 A

A capacitor can be used in circuits with induction coils to smooth out voltage spikes and reduce arcing at the contacts. It helps absorb sudden changes in voltage, thereby minimizing wear and prolonging the life of the contacts.

The inductive reactance (X\(_L\)​) of an inductor is given by the formula: Given: X\(_L\) = 1\(\Omega\), f = 50Hz X\(_L\) = 2\pifL L = \(\frac{1}{2\pif}\) = \(\frac{1}{2\pi \times50}\) = \(\frac{1}{100\pi}\) = 3.183 x 10\(^{-3}\)H

\(\frac{1}{\text{R}}\) = \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{5}{6}\) R = \(\frac{6}{5}\) = 1.2\(\Omega\); So p.d through the 3\(\Omega\) and 2\(\Omega\) resistors in parallel = 1.5 x 1.2 =1.8V So the current through the 3\(\Omega\) resistor(I) = \(\frac{\text{V}}{\text{R}}\) = \(\frac{1.8}{3}\) = 0.6A (recall V = IR)

"magnetic field" is the term that describes the region where magnetic effects are felt.

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

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

To calculate the peak voltage (\(V_p\)) from the root mean square (RMS) voltage (\(V_{rms}\)), we use the formula: \(V_p = V_{rms} \times \sqrt{2}\) Given: \(V_{rms} = 220 \, \text{V}\) Calculate the Peak Voltage \(V_p = 220 \, \text{V} \times \sqrt{2} \approx 220 \, \text{V} \times 1.4142 \approx 311.1 \, \text{V}\) The peak voltage is approximately: 311 V

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

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

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