ACTIVITY 1

Example Exercise 1 Finding the current by Ohmfs law. A 6.00-V battery is Find the resistance by Ohmfs law. A potential applied to a circuit having a resistance of 10.0 (2. Find difference of 12.0 V is applied to a circuit and a the current in the circuit. current of 0.300 A is observed. Solution: The current in the circuit, found by Ohm's law. V 6.00 V V E 10.09 0.60? 0.600A A Exercise 2 Exercise 3 What value of resistance is necessary to get a current Find the current by Ohms law. A 500.0 Q resistor is of 2.50 A when it is connected to a 120 V source? connected to a 12.0 V battery. Find the current through the resistor Example Exercise 4 The resistance of a spool of wire. Find the resistance The resistance of an amber rod. Find the resistance of of a spool of copper wire, 500 m long with a diameter an amber rod 30.0 cm long by 2.00 cm high and d of 0.644 mm. The p of copper is 1.72 x 1089 m 2.50 cm think. The p of amber is 5.00 x 10149 In Solution Solution The cross-sectional area of the wire is given by The resistance from one end of the rod to the other is no!2 ' T 4 2 6.44 10' = W = 3.26 x 10-7m2 The resistance of the wire spool, is l R = pg (500 m) R = (1.72 x 1089 m) R = 26.4 0 (3.26 x 10'7m2) Exercise 5 Exercise 6 A water heater draws 30 A from a 120.0 V power The resistance of a spool of copper wire. Find the source 10 m. away. What is the minimum cross resistance of a 100.0 m spool of #22 gauge copper section of the wire that can be used if the voltage wire (diameter d = 6.44 x 10'4m) The p of copper is applied to the heater is not lower than 115.0 V? The 1.72 x 1039 m permissible voltage drop is 5.0 V, and the resistance A _ Hdz _ 1T(ii-449610"4 m)2 that corresponds to this drop when the current is _ T _ 4 30 A is A = 3.26 x 10'7m2 V 5.0 V Solution R = T = m = 0'167 Q The resistance from one end of the rod to the other is The total length of wire involved is twice the distance between the source and the heater so L = (2) (10 m) = 20m. The p of copper is 1.72 x 1039 m l l |R=PIFA=PE Exercise 7 The resistance of a silver wire. A silver wire 2.0 m long is to have a resistance 0.50 9. What should its diameter be? The p of silver is 1.59 x 1089 m Solution The resistance from one end of the rod to the other is Example Exercise 8 Temperature dependence of resistivity. The resistivity Temperature dependence of resistivity. The of copper at 20.0 C is 1.72 x 10-80 m. Find its resistivity of gold at 20.0 C is 2.44 x 10-80 m. Find resistivity at 200.0 .C its resistivity at 100.0 .C Solution Solution The temperature coefficient of resistivity for copper is The temperature coefficient of resistivity for gold is 3.93x10 .C The resistance of copper resistor at 200.0 C is 3.40x10- .C The resistance of gold resistor at 100.0 C is p = Po[1 + a(t - to)] 3.93x10-3 p = 1.72 x 10-80 m 1+ (200.0 - 20.0) .C p = 2.94 x 10-80m Exercise 9 Temperature dependence of resistivity. The resistivity of iron at 20.0 C is 9.71 x 10 802 m. Find its resistivity at 100.0 .C Solution The temperature coefficient of resistivity for iron is 5.20x10 -3 The resistance of gold resistor at 100.0 C is .CExample Exercise 10 Temperature dependence of resistance. If the The resistance of a resistor is 20.0 2 at 20.0 C. Find resistance of a copper resistor is 50.0 2 at 20.0 C, find its resistance at 200.0 C. The temperature coefficient its resistance at 200.0 .C of resistance a for copper is 3.93x10-3 .C Solution The temperature coefficient of resistivity for copper is 3.93x10-3 The resistance of copper resistor at 200.0 .C is .C R = Ro[1 + a(t - to)] 3.93x10-3 R = (50.0 0) 1+ .C (200.0 - 20.0) .C R = 85.4 Q Exercise 11 Exercise 12 If the resistance of a copper wire is 500.0 2 at 20.0 .C. Suppose the resistance of a copper conductor is Find its resistance at 100.0 C. The temperature 1.72 Q at a temperature of 20.0 C. Find its resistance coefficient of resistance a for copper is 3.93x10-3 at 0.0 .C and 100.0 C. The temperature coefficient of .C resistivity for copper is 2 $ 3.93x10-3 .CExample A JZU-W/tht bulb. What is the current and resistance of a 120W light bulb connected to a 120 V source? Solution P=IV;I= V=IR; R P _ (120.0 W) v _ (120.0V) _ _ V _ (120.0 V) _ 1' (1.00A) _ AV 1.00 = 1.00A V 1209 Exercise 13, 14, 15 A 50-W/tht bulb. A 60.0 W light bulb is screwed into a 120.0 V lamp outlet. What current ow through the bulb and what is the resistance of the bulb? What is the power dissipated in the Joule heating of the wire in the bulb? Solution How to calculate electric usage cost. Electricity usage is calculated in kilowatt-hours. A kilowatt-hour is 1,000 watts used for one hour. As an example, a 100-watt light bulb operating for ten hours would use one kilowatt-hour. 1. Electric Power = Current x Voltage Watts 1000 = Kilowatts (KW) 2. Energy = power x time Kilowatt Hours = Kilowatts (kW) x Hours of Use 3. Cost of Usage = Kilowatt Hours (kWh) x kWh rate 4. Add other fixed metering and distribution tax charges Where electrical energy costs PhP 9.00 per kilowatt- How much power is used by a calculator that hour, Compute the daily electrical usage cost of a 100- operates on 8 volts and 0.1 ampere? If it is used for watt light bulb that has been used for 10 hours, and if 8 hours, how much energy does it use? What is the used daily for 30 days cost of usage? 1. Kilowatts = 100 W = 0.10 kW 1. Power = current x voltage = (0.1 A) x (8 V) = 1000 2. Energy = 0.10 kW x 10 h = 1 kWh 0.8 W. 3. Cost of usage = 1 kWh x PhP 9.00 2. Energy = power x time = (0.8 W) x (8 h) = PhP 9.00 kWh 6.4 Wh = Php 9.00 Kilowatts = 0.0064 kWh 4. Monthly usage = = x 30 days = PhP 270.00 1000 day 3. Cost of usage = 0.0064 kWh x PhP 9.00 = PhP 0.057 kWh A 2,000 W bread toaster is being used for 15 minutes PhP 9.00 in the morning to toast bread slices for breakfast. How Cost of usage = 0.5 kWh x = PhP 4.50 kWh much is the monthly usage of the unit? PhP 4.50 2000 W Monthly usage = x 30 days = PhP 135.00 day Kilowatts = = 2.0 kW 1000 Energy = 2.0 kW x 0.25 h = 1 kWh