Electricity Electric field Electric field strength Warning: β = Ε Energy and electric field strength use the same variable. πΉ π β: Ε ββββ πΉπΈ : π: Electric field strength (π π−1 ) electric force (π) electric charge (πΆ) (Coulomb Field strength between two parallel plates Ε= π π π: π: Voltage (π) distance (π ) AC vs DC There are two types of electric current • DC, direct current, where electrons flow in one direction • AC, alternating current, where the direction of the electron flow is reversed periodically. The following laws can be applied to both currents. Electric circuit 1 • There is a to be a voltage potential. • Current will flow only when circuit is closed. • Energy is transported by electric field, not electrons. • Normally the resistance of the wires is neglected. Ohms law π = π πΌ π: voltage(π) [Volt] π : resistance (πΊ) [Ohm] Voltage π= π π π: voltage (π ) [Volt] π: electric charge (πΆ ) [Coulomb] Current πΌ= π π‘ πΌ: current (π΄) [Ampere] π‘: time (π ) [secomds] 2 Power π = ππΌ π: power(π) [Watt] Energy πΈ = ππ‘ πΈ: Energy (π½) [J] Work π = πΉπ π: work (π½) [Joule] πΉ: force (π) [Newton] π : distance (π) [metre] Kinetic energy πΈπ = 1 ππ£ 2 2 πΈπ = π πΈπ : kinetic energy (π½) π: mass (ππ) π£: velocity (ππ −1 ) ] Kirchhoff's laws Current law πΌin = πΌππ’π‘ πΌ1 + πΌ2 + πΌ3 = πΌ4 + πΌ5 + πΌ6 3 Voltage law ππ = π1 + π2 + π3 + π4 + π5 The resistance of the wires is neglected. Series circuit πtotal = π1 + π2 + β― + ππ π π‘ππ‘ππ = π 1 + π 2 + β― + π π πΌ= ππ‘ππ‘ππ π π‘ππ‘ππ Example (diagram) π = (π 1 + π 2 )πΌ 4 Parallel circuit π = π π‘ππ‘ππ πΌπ‘ππ‘ππ 1 π π‘ππ‘ππ = 1 1 1 + + β―+ π 1 π 2 π π πΌtotal = πΌ1 + πΌ2 + β― πΌπ Example (diagram) π= π 1 π 2 πΌ π 1 + π 2 5 Problems Question 1 A battery supplies 3.50 (π΄) to operate an electric motor for 30 (min). Calculate the total amount of charge that flowed from the battery. Question 2 A battery supplies 200(πΆ) in 3 (min). Calculate the electric current Question 3 An appliance is rated at 2.40 (ππ ) when a current of 10.0 (π΄) is flowing through it. If the appliance is operated for 1 (β) determine: a) The potential difference across the appliance b) The amount of heat converted from electric energy c) Total charge flowing through the appliance Question 4 An electric circuit composed of a 4.5 (π) battery a switch and a 2.50(π) light bulb. The switch is closed for 30 (π ). Calculate: a) The current flowing through the circuit b) The quantity of charge flowing through the circuit. c) The electrical energy converted by the globe to light and heat. Question 5 A small electric motor requires a 3 (π ) battery. If the motor draws a maximum current of 300 (ππ΄) calculate: a) The resistance of motor. b) The maximum power consumption 6 Question 6 A string of 16 light globes is connected in series to a mains outlet of 240 (π ). If the total power consumption is 24.0 (π), calculate: a) the potential difference across each light globe b) the resistance of each light globe Question 7 For the circuit above calculate a) The total resistance of the circuit b) The current πΌ Question 8 For the circuit above calculate a) The total resistance of the circuit b) The current πΌ c) The current πΌ1 and πΌ2 7 Question 9 For the circuit shown calculate: a) The total resistance b) The current πΌ. c) The potential difference across the 24 (Ω) resistor d) The power dissipated through the 8 (Ω) resistor. Question 10 The above circuit has a rheostat, which is a variable resistor. The rheostat was adjusted for a current of 0.5 (π΄). Calculate a) the value of π that the voltmeter will show. b) the resistance of the rheostat on this setting. 8 Question 11 For each of the following re-draw the diagram and draw the electric field distribution associated with is a) b) c) d) e) 9 Solutions Question 1 A battery supplies 3.50 (π΄) to operate an electric motor for 30 (min). Calculate the total amount of charge that flowed from the battery. πΌ= π π‘ π = πΌπ‘ π = 3.50 ∗ (30 ∗ 60), 1 (min) = 60(π ) π = 6 300 (πΆ) Question 2 A battery supplies 200(πΆ) in 3 (min). Calculate the electric current πΌ= π π‘ πΌ= 200 3 ∗ 60 πΌ = 1.11 (π΄) Question 3 An appliance is rated at 2.40 (ππ ) when a current of 10.0 (π΄) is flowing through it. If the appliance is operated for 1 (β) determine: a) The potential difference across the appliance π = ππΌ π= π πΌ π= 2 400 10 π = 240 (π) 10 b) The amount of heat converted from electric energy Heat is thermal energy. Electric energy is converted to heat energy. πΈ = ππ‘ πΈ = (2.4 ∗ 103 ) ∗ 3600, 1(β) = 3600 (π ) πΈ = 8.64 ∗ 106 (π½) c) Total charge flowing through the appliance π = πΌπ‘ π = 10.0 ∗ 3600 π = 3.6 ∗ 104 (πΆ) Question 4 An electric circuit composed of a 4.5 (π) battery a switch and a 2.50(π) light bulb. The switch is closed for 30 (π ). Calculate: a) The current flowing through the circuit π = ππΌ πΌ= π π πΌ= 2.50 4.5 πΌ = 0.55 (π΄) 11 b) The quantity of charge flowing through the circuit. π = πΌπ‘ π = 0.55 ∗ 30 π = 16.7 (πΆ) c) The electrical energy converted by the globe to light and heat. πΈ = ππ‘ πΈ = 2.50 ∗ 30 πΈ = 75 (π½) Question 5 A small electric motor requires a 3 (π ) battery. If the motor draws a maximum current of 300 (ππ΄) calculate: a) The resistance of motor. π πΌ 3 π = 0.3 π = π = 10 (Ω) b) The maximum power consumption π = ππΌ π = 3 ∗ 0.3 π = 0.9 (π) 12 Question 6 A string of 16 light globes is connected in series to a mains outlet of 240 (π ). If the total power consumption is 24.0 (π), calculate: a) the potential difference across each light globe Series circuit ππΊππππ = ππ‘ππ‘ππ πππππππ 240 16 = 15 (π) ππππππ = ππππππ b) the resistance of each light globe 1. Resistance of circuit π = ππΌ, πΌ= 2. Resistance of one globe π π π πππππ = π2 π= π π2 π = π 2402 π = 24 2400 16 = 150 (Ω) π πππππ = π πππππ π = 2400 (Ω) 13 π π‘ππ‘ππ ππππππ Question 7 For the circuit above calculate a) The total resistance of the circuit Series circuit π π‘ππ‘ππ = π 1 + π 2 π π‘ππ‘ππ = 11 + 7 π π‘ππ‘ππ = 18 (Ω) b) The current πΌ π π 9 πΌ= 18 πΌ= πΌ = 0.5 (π΄) 14 Question 8 For the circuit above calculate a) The total resistance of the circuit Parallel circuit 1 = 1 1 + π 1 π 2 π π‘ππ‘ππ = π 1 π 2 π 1 + π 2 π π‘ππ‘ππ = 6 ∗ 18 6 + 18 π π‘ππ‘ππ π π‘ππ‘ππ = 4.5 (Ω) b) The current πΌ πΌ= πΌ= π π π‘ππ‘ππ 9.0 4.5 πΌ = 2 (π΄) 15 c) The current πΌ1 and πΌ2 1. π°π 2. π°π π π 9.0 πΌ= 18 π π 9 πΌ= 6 πΌ= πΌ= πΌ = 0.5 (π΄) πΌ = 1.5 (π΄) Question 9 For the circuit shown calculate: a) The total resistance 1. Resistance in parallel branch π π‘ππ‘ππ = π π‘ππ‘ππ π π‘ππ‘ππ 2. Total resistance, series π 1 π 2 π 1 + π 2 π π‘ππ‘ππ = π 1 + π 2 π π‘ππ‘ππ = 8 + 16 48 ∗ 24 = 48 + 24 = 16 (Ω) π π‘ππ‘ππ = 24 (Ω) b) The current πΌ. π π 6.0 πΌ= 24 πΌ= πΌ = 0.25 (π΄) 16 c) The potential difference across the 24 (Ω) resistor Parallel circuit the voltage drop is determined for all resistors in the circuit π = π π‘ππ‘ππ πΌπ‘ππ‘ππ π = 16 ∗ 0.25 π = 4 (π) d) The power dissipated through the 8 (Ω) resistor. π = ππΌ, π = π πΌ π = π πΌ 2 π = 0.5 (π) Question 10 The above circuit has a rheostat, which is a variable resistor. The rheostat was adjusted for a current of 0.5 (π΄). Calculate a) the value of π that the voltmeter will show. In a series circuit the current is constant across the circuit. π = π πΌ π = 20 ∗ 0.5 π = 10 (π) 17 b) the resistance of the rheostat on this setting. 1. Voltage drop across Rheostat 2. Resistance of Rheostat ππ βππ = ππΆπππ − ππ ππ ππ π‘ππ π = π πΌ 2 π = 0.5 π = 12 − 10 π = 2 (π) π = 4(Ω) Question 11 For each of the following re-draw the diagram and draw the electric field distribution associated with is a) b) c) d) 18 e) 19
