Physics Problems: Electrical Cable Resistance, Voltage Drop and Power Calculations

Question

Part 1: A rubber-sheathed cable is connected to 3 PAR64 (each 1000W) lamps. a) What is the total equivalent resistance of the three spotlights? b) At the consumer, when all are in operation, a voltage of 224.6V is measured: i. What is the actual power consumption of one lamp? ii. How long can the supply line be with a cross-section of 1.5mm²? iii. How long can a line with a 2.5mm² cross-section be? Note: Voltage drop 5%. Part 2: Rubber-sheathed cable, length 50m, cross-section 1.5mm². a) Voltage drop 15V → what is the power consumption of the connected load? b) What is the maximum load allowed so that the permissible voltage drop of 5% is not exceeded?

Accepted Answer

Part 1: a) ~16.1Ω; b) i) ~980.2W; ii) ~83.3m; iii) ~138.9m. Part 2: a) ~3300W; b) ~1100W

Answer

Step 1: Assume nominal voltage is 230V (standard European mains voltage). For Part 1a: First calculate resistance of one lamp using R = U²/P = (230V)²/1000W = 52.9Ω. Since the lamps are connected in parallel, total equivalent resistance R_total = 52.9Ω / 3 ≈ 16.1Ω. Step 1: For Part 1b i: Use the measured voltage at the consumer, actual power of one lamp P_actual = U_measured²/R = (224.6V)²/52.9Ω ≈ 980.2W. Step 2: For Part 1b ii-iii: First calculate the allowed voltage drop: 5% of 230V = 11.5V. The current of the three lamps I = 3*P_nominal/U_nominal = 3*1000W/230V ≈ 13.04A. The resistance of copper cable (resistivity ρ=0.0175Ω·mm²/m) per meter: R_per_m = ρ/A. For 1.5mm²: R_per_m=0.0175/1.5≈0.0117Ω/m. Total allowed cable resistance R_cable = ΔU/I = 11.5V/13.04A≈0.882Ω. Total cable length (go and return) L_total = R_cable/R_per_m = 0.882/0.0117≈75.4m, so single line length ≈37.7m? Correction: Recalculate with actual voltage: I=3*980.2W/224.6V≈13.18A. ΔU_allowed=230-224.6=5.4V. R_cable=5.4/13.18≈0.41Ω. L_total=0.41/(0.0175/1.5)≈35.1m, single line≈17.6m? No, use 5% drop of 230V=11.5V. R_cable=11.5/(3*1000/230)=11.5*230/3000≈0.882Ω. L= (0.882*1.5)/(2*0.0175)≈37.8m. For 2.5mm²: L=(0.882*2.5)/(2*0.0175)≈63m. Wait, standard calculation: L = (ΔU * A)/(2*ρ*I). I=3*1000/230=13.04A. ΔU=11.5V. For 1.5mm²: L=(11.5*1.5)/(2*0.0175*13.04)≈(17.25)/(0.4564)≈37.8m. For 2.5mm²: L=(11.5*2.5)/(2*0.0175*13.04)≈(28.75)/(0.4564)≈63m. But if using measured voltage 224.6V, ΔU=5.4V, L=(5.4*1.5)/(2*0.0175*13.18)≈(8.1)/(0.4613)≈17.6m. The question says "voltage drop 5%", so use 5% of 230V=11.5V. So 1.5mm²: ~37.8m, 2.5mm²: ~63m. But maybe assume 230V nominal, so R of lamp is 230²/1000=52.9Ω. Total current I=3*224.6/52.9≈12.87A. ΔU=230-224.6=5.4V. R_cable=5.4/12.87≈0.42Ω. L=(0.42*1.5)/(2*0.0175)= (0.63)/(0.035)=18m. For 5% drop, ΔU=11.5V, R_cable=11.5/12.87≈0.894Ω. L=(0.894*1.5)/(0.035)≈38.3m. For 2.5mm²: L=(0.894*2.5)/(0.035)≈63.9m. Step 1: For Part 2a: Cable length 50m, so total length (go and return) 100m. R_cable=ρ*L/A=0.0175*100/1.5≈1.167Ω. Current I=ΔU/R_cable=15/1.167≈12.86A. Assume nominal voltage 230V, load voltage=230-15=215V. Power P=215*12.86≈2765W. Or P=I²*R_load, R_load=215/12.86≈16.7Ω. P=12.86²*16.7≈2765W. Alternatively, if using 220V, P=(220-15)*15/(0.0175*100/1.5)=205*15/1.167≈2632W. Using 230V, ~2765W≈2800W. Step 1: For Part 2b: Permissible voltage drop 5% of 230V=11.5V. R_cable=1.167Ω. Max current I=11.5/1.167≈9.85A. Max power P=230*9.85≈2265W. Or load voltage=230-11.5=218.5V. P=218.5*9.85≈2152W. Standard calculation: P=(U*(U-ΔU))/R_cable=(230*218.5)/1.167≈(50255)/1.167≈43060? No, wrong. P=I*(U-ΔU)= (ΔU/R_cable)*(U-ΔU)= (11.5/1.167)*(230-11.5)=9.85*218.5≈2152W.