Question

relatively straight forward, please show all work!

2) Transmission losses Imagine that you want to provide power to a resistive element (a load) with a battery as shown in the diagram below. Your load needs 20W of power to operate. You have two choices: you can connect a 5V or a 12 V battery. a) How much current must run through the circuit in each case? b) The wiring in your dorm uses a standard 12 gauge wire size with a resistance of around 5.2 x 10-12/meter. If you use 10 meters of such wire to make your circuit how much power will you lose to heating the wire in each case? c) Now think about Bowdoin's ground-mounted solar array. It's around 2 km from campus. The peak power capacity is 654 kW. Let's assume that the panel connections are managed so that the DC voltage from the whole array is 12 kV. Instead of being transmitted down Harpswell Road at this voltage, the voltage is converted to an AC voltage and stepped up to 120 kV for transmission. For these uses, we need to use a larger 6 gauge transmission wire with a resistance of 1.3 x 10" 22/meter. How much power will you lose in the wire transmission if the energy is transmitted at 12 kV rather than 120 kV? How much worse would the losses be if instead of transforming the voltage up to 120 kV, you transformed it down to 120 V (that is, you transmitted the power at the voltages used in outlet wiring)?

Answer 1

Solution : Power required = 20W V Vhigh Resistive element @ if battery is of You T Battery a Voltage 5V - 5V, then, using, VP =P a PE VI, cument, We have, Current, I = ŕ = 20W = 4A Answer if battery is 12v, then, Current, 1= = 20W - A = 1.67A Answer 6. Resistance of the Tom e wire.is, R = 5.2 x 1032/meter x 10 meters : = 5.2 xlo? 2 Now, for sv Battery, Power dissipated is, P = x2 - Com 25 10° wet = 480.8 W ~ 481 W for 12V Battery, Power dissipated in healing - the wire is, B v2 (2v)__ = 2769 W = 2.8 kW cs) Scat 2. T WIĆ 528102e 1 And CamScanner

(0) Power to be transmitted = 654 kW = Rllet say) and length of wire used = 2km = 2000m (it's the distance 2km from Campus ) Further, the resistance per unit length is 1.381032/meter. i total resistance of the wire used for transmission = R = 1.3x10° 21m x 2000m → R = 2.612 Now, if Power is transmitted at 12kV, then, - Current in conducter, I = P 6 54 KW 12 kV = 54.5 A Therefore, Power 1088 = 1?R = 54.5)2 (2.6) W a = 7722.6W = 7.7 kW if Power is transmitted at 120 kv, then; Str.120 KV I= P - 654 KW = 5.45A Hence, Power loss = 12R = (5-45 (216) W = 77.2W R 0.077 kW Further, if Power is transmitted . at 120V, then, LES Scanned wit1 = PO = 654 kW = 5450 A CamScanner v

So, Power loss in this case is, IPR = (54503(2.6) W = 77226500 W = 77.2 MW (huge Power loss!) Power loss ratio at( 120 V to 120KV) is, 77226500 26509 = 106 times ! 77.2 Very Worse ! so. cs Scanned with CamScanner