Flashcards
Lectures 1 & 2
Cards in this deck (27)
- 1What is 1 Coulomb equal to?6.242 x 10 ^ 18
- 2Given time and charge, how would you calculate the current?I = Q/t
Where I = Current
Where Q = Charge
Where t = Time in seconds - 3Given the current and time, how would you calculate the charge?Q = It
Where I = Current
Where Q = Charge
Where t = Time in seconds - 4Given the charge and current, how would you calculate the time?t = Q/I
Where I = Current
Where Q = Charge
Where t = Time in seconds - 5The charge flowing through a point in a copper conductor is 0.16 C every 64 ms. Determine the current in amperes?I = Q/t
I = 0.16/ 64ms
I = 0.16/ .064
I = 2.5A - 6Determine the time required for 4 x 10 ^16 electrons to pass through a point in a copper wire if the current is 5 mA.t = Q/I
4 x 10 ^16/ 5mA
Electrons to Coulombs = 4x 10 ^16 / 6.242 x 10 ^18 = .0064 C
.0064/5mA = .0064/.005
t = 1.28 s (seconds) - 7What is the formula for potential difference when given energy (W) and charge (Q)?V = W/Q volts
Where V = Voltage
Where Q = Charge
Where W = Expended Energy - 8What is the formula for potential difference when given voltage (V) and charge (Q)?W = QV Joules
Where V = Voltage
Where Q = Charge
Where W = Expended Energy - 9What is the formula for potential difference when given expended energy (W) and voltage (V)?Q = W/V Coulombs
Where V = Voltage
Where Q = Charge
Where W = Expended Energy - 10Find the potential difference between two points in an electrical system if 60 J of energy are expended by a charge of 20 C between these two points.V = W/Q
V = 60/20
V = 3 V - 11Determine the energy expended moving a charge of 50 uC through a potential difference of 6V.W = QV Joules
W = 50 x 10 ^-6 * 6
W = 30 x 10 ^ -6 J
Acceptable answers are: 30 x 10^-6J or 300uJ or .3mJ or .0003J - 12Draw the circuit diagram for resistance and what is it measured in?
- 13State Ohms law?Ohms law = V = IR
Where V = Voltage
Where I =Current
Where R = Resistance - 14How would you calculate the Voltage given the Current and Resistance?Ohms law = V = IR
Where V = Voltage
Where I =Current
Where R = Resistance - 15How would you calculate the Current give the Voltage and Resistance?Ohms law = I = V/R
Where V = Voltage
Where I =Current
Where R = Resistance - 16How would you calculate the Resistance given the current and Voltage?Ohms law = R = V/I
Where V = Voltage
Where I =Current
Where R = Resistance - 17A current of 5.5mA flows through a 33 k resistor. Calculate the p.d. thus developed across it.V =IR
V = 5.5 x 10^ -3 * 33 x 10^3
V = 181.5V - 18If a p.d. of 24V exists across a 15 ohm resistor then what must be the current flowing through it?I = V/R
I = 24/15
I = 1.6A - 19How would you calculate energy dissipated given Current, Resistance and time?W = I^2Rt
Where I = Current
Where R = Resistance
Where t = Time - 20How would you calculate energy dissipated given Voltage, Resistance and time?W = V^2t/R
Where V = Voltage
Where R = Resistance
Where t = Time - 21How would you calculate energy dissipated given Voltage, Current and time?W = VIt
Where V = Voltage
Where I = Current
Where t = Time - 22A current of 200mA flows through a resistance of 750 ohms for a time of 5 minutes. Calculate (a) the p.d. developed, and (b) the energy dissipated.(a) W = i^2Rt => (200x10^3)^2 x 750 x 5 x 60
W = 9000J
(b) W = VIt so V = W/It
9000/(200x10^-3) * 5 *60
V = 150V - 23How can we derive power and what is it measured in?Power is Energy divided by time and the Measurement is in Watts (W)
P = W/t
Where W is Energy
Where t is Time - 24Given the Current, Resistance and time, what formula would you use to calculate Power?W = I^2Rt
P = I^2R
Where I = Current
Where R = Resistance
Where W = Energy
Where t = time in seconds - 25Given the Voltage, Resistance and time, what formula would you use to calculate Power?W = V^2t/R
P = V^2/R
Where I = Current
Where R = Resistance
Where W = Energy
Where t = time in seconds - 26Given the Voltage, Current and time, what formula would you use to calculate Power?W = VIt
P = VI
Where I = Current
Where W = Energy
Where t = time in seconds - 27A current of 1.4 A when flowing through a circuit for 15 minutes dissipates 200 kJ of energy. Calculate (a) the p.d., (b) power dissipated and (c) the resistance of the circuit(a) W = I^2Rt => R = W/I^2t
R = 200x10^3/1.4^2 * 15 *60 = 113 ohms
(b) P = I^2R = 1.4^2 * 113 = 222 W
(c) P = VI => V = P/I
222/1.4 = 158 V
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- NVQ Level 6 Instrumentation Flashcards on Lectures 1 & 2, created by carolinepgray on 23/09/2013.
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