This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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d) For the circuit, calculate: i. The total circuit resistance Step 1: Calculate the equivalent resistance of the three 10Ω resistors in parallel. (1)/(R_p) = (1)/(10\,) + (1)/(10\,) + (1)/(10\,) (1)/(R_p) = (3)/(10\,) R_p = (10)/(3)\, Step 2: Calculate the total circuit resistance by adding the parallel equivalent resistance to the series resistors (5Ω and 10Ω). R_total = R_p + 5\, + 10\, R_total = (10)/(3)\, + 5\, + 10\, R_total = (10)/(3)\, + (15)/(3)\, + (30)/(3)\, R_total = (55)/(3)\, The total circuit resistance is (55)/(3)\, or 18.33\, ii. The voltage-drop across the 5Ω resistor Step 1: Calculate the total current flowing from the 10V supply using Ohm's Law (V = IR_total). I_total = V_supplyR_total I_total = 10\,V(55)/(3)\, I_total = (30)/(55)\,A = (6)/(11)\,A Step 2: Calculate the voltage drop across the 5Ω resistor using Ohm's Law (V = IR). Since the 5Ω resistor is in series, the total current flows through it. V_5 = I_total × 5\, V_5 = (6)/(11)\,A × 5\, V_5 = (30)/(11)\,V The voltage drop across the 5Ω resistor is (30)/(11)\,V or 2.73\,V e) List any four sources of electricity. • Batteries (chemical energy) • Generators (mechanical energy) • Solar cells (light energy) • Nuclear power plants (nuclear energy) 12. a) State the SI units of the following electrical quantities: i. Conductance: Siemens (S) ii. Inductance: Henry (H) iii. Charge: Coulomb (C) iv. Resistivity: Ohm-meter (Ω·m) v. Capacitance: Farad (F) b) A 60 W security bulb operates on 240V i. Find the current flowing through it. Step 1: Use the power formula P = VI. I = (P)/(V) I = 60\,W240\,V I = (1)/(4)\,A The current flowing through the bulb is 0.25\,A ii. Calculate the energy it consumes overnight in kilojoules if it is left on for 12 hours. Step 1: Convert the time from hours to seconds. t = 12\,hours × 3600\,secondshour = 43200\,s Step 2: Calculate the energy consumed using the formula E = P × t. E = 60\,W × 43200\,s E = 2592000\,J Step 3: Convert the energy from Joules to kilojoules. E = 2592000\,J1000 E = 2592\,kJ The energy consumed is 2592\,kJ c) Two bulbs of 100W and 40W respectively are connected in series across a 240V supply. Which bulb will glow brighter and why? The 40W bulb will glow brighter. When bulbs are connected in series, the same current flows through both. The brightness of a bulb depends on the power it dissipates, which is given by P = I^2R. The resistance of a bulb is inversely proportional to its rated power (R = V_rated^2/P_rated). Therefore, the 40W bulb has a higher resistance (R_40W = (240\,V)^2 / 40\,W = 1440\,) compared to the 100W bulb (R_100W = (240\,V)^2 / 100\,W = 576\,). Since the current (I) is the same for both, the bulb with the higher resistance will dissipate more power (I^2R) and thus glow brighter. d) Differentiate between a motor and a generator. A motor converts electrical energy into mechanical energy, causing motion. A generator converts mechanical energy into electrical energy, producing an electric current. e) State Coulomb's Law. Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges. Send me the next one 📸