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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.

X\(_L\) = 2\(\pi\)fL L = 7H, f = 50Hz X\(_L\) = 2 x 3.142 x 7 x 50 = 2199.40 \(\approxeq\) 2200\(\Omega\)

Impedance is a measure of how much a circuit opposes the flow of alternating current (AC). It combines both resistance (the real part) and reactance (the imaginary part) of a circuit component or network. Impedance is denoted by the symbol and is measured in ohms (\Omega).

At resonance, the inductive reactance (V\(_L\) ) equals the capacitive reactance (V\(_C\)), causing the voltages across the inductor and capacitor to be equal in magnitude but opposite in phase.

Effective voltage in an RLC circuit is given by V = \(\sqrt{V_R^2 + (V_L - V_C)^2}\) V = \(\sqrt{3^2 + (6 - 2)^2}\) = \(\sqrt{3^2 + 4^2}\) = \(\sqrt{9 + 16}\) = 5V

X\(_c\) = \(\frac{1}{2\pi FC}\) X\(_c\) = \(\frac{1}{2 \times 3.142 \times 60 \times 2 \times 10^{-6}}\) X\(_c\) = 1326.1192\(\Omega\) I = \(\frac{\text{V}}{X_c}\) I = \(\frac{150}{1326.1192}\) = 0.113A

To calculate the energy of light, we can use the formula: E = h x f h = 6.63 x 10\(^{-34}\)Js f = 2.0 x 10\(^{15}\)Hz E = 6.63 x 10\(^{-34}\) x 2.0 x 10\(^{15}\) = 1.326 x 10\(^{-18}\)J

Energy of the emitted photon(E) = \(\frac{\text{hc}}{\lambda}\) \(\lambda\) = 5.68 x 10\(^{-6}\)m h = 6.63 x 10\(^{-34}\)Js c = 3.0 x 10\(^8\)ms\(^{-1}\) E =\(\frac{(6.63\times 10^{-34}) \times (3.0 x 10^8)}{5.68\times 10^{-6}}\) = \(\frac{1.989\times 10^{-25}}{5.68\times 10^{-6}}\) E = 3.50 x 10\(^{-20}\)J Therefore, the energy change = 3.50 x 10\(^{-20}\)J

Infrared thermometers work by detecting the infrared radiation emitted by objects. They measure this radiation to determine the object's temperature without direct contact.

A photometer is used to measure the intensity of light. Photometers can determine the brightness of light sources, compare light intensity from different sources, or measure the amount of light reaching a specific point. They are commonly used in photography, astronomy, and various scientific applications to ensure proper lighting conditions

Cathode rays are streams of electrons emitted from the cathode (negative electrode) in a vacuum tube. Cathode rays can be deflected by magnetic and electric fields, demonstrating their charge and nature as electrons.

E = hf + W\(_o\) E = 1.12 x 10\(^{-18}\)J Max. Energy = 8.0 x 10\(^{-19}\)J. W\(_o\) = E - Max. Energy = 1.12 x 10\(^{-18}\)J - 8.0 x 10\(^{-19}\)J. = 3.2 x 10\(^{-19}\)J. But W\(_o\) = hf\(_o\) f\(_o\) = \(\frac{W_o}{h}\) = \(\frac{3.2 \times 10^{-19}}{6.626 \times 10^{-34}}\) = 4.83 x 10\(^{14}\)Hz

(a) electrons are only emitted when the photon frequency exceeds the threshold frequency, - the energy of each photon is proportional to its frequency, - the emission of photoelectrons depends on the frequency of incident light on the metal, - the intensity of light does not affect the energy of the electrons emitted/photoelectrons, - the energy of the photoelectron depends on the frequency of incident radiation. (b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ? Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\) = \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\) = 6 x 10\(^{-19}\)J But K.E = E + W\(_o\) W\(_o\) = E - K.E = E - eV W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J (c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc (d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity. But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\) \(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\) Recall, A = \(\lambda\)N N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei