Electric Circuits Flashcards
Define P.D
Energy transferred per unit charge (V = E/Q)
Define EMF
Work done per unit charge by the power supply (converting chemical energy into electrical potential energy of the charges)
Define Current
Rate of flow of charge
Define resistance
Ratio of P.D to current
P.D divider derivation
- I is same
- V1 + V2 + V3 = VT
- I = V/R
- VT/RT = V1/R1 + V2/R2
Parallel current derivation
- V is same
- V = IR
- I1R1 = I2R2
Series resistor derivation
- I is same
- V1 + V2 + V3 = VT
- I = V/R
- RT = R1 + R2 + R3
Parallel resistor derivation
- I is same
- V1 + V2 + V3 = VT
- I = V/R
- 1/RT = 1/R1 + 1/R2 + 1/R3
P.D law
Sum of the P.D drops = sum of the emf’s around a closed loop within a circuit (conservation of energy)
Current Law
Sum of currents into a junction = sum of the currents out of the junction (conservation of charge)
EMF & Ir circuit
- ammeter in series
- voltmeter parallel to a variable resistor
- (draw cell resistor)
Define lost volts
Energy per unit charge transferred to the internal resistance of a cell
Define power
Rate of energy transfer (P = E/t)
Derive P = I2R & P = V2/R
- P = IV
- I = V/R
- P = V2/R
- P = IV
- V = IR
- P = I2R
Brighter bulb in series (B) or parallel (A)
*P.D in each bulb in A is equal
*P = V2/R, lower resistance bulb has higher power
*So 20Ω bulb is brighter in A
*I in each bulb in B is equal
*Therefore, as P = I2R, the bulb with the higher resistance will have a higher power
*So the 40Ω bulb is brighter in B
no. electrons in a certain amount of charge calculation
- Number of electrons = total charge / charge on one electron (n = Q/e)
- N = Q/ 1.6 x10 -19
I = nqvA
I = current
n = number density of free charge carriers (per unit vol.)
q/e = charge on 1 electron
v = (average) drift vel.
A = cross sectional area of current carrying material
Derive I = nqvA
- There are N electrons moving with a drift velocity v along a wire of length l and cross sectional area A
- The current I = ∆Q/∆t
- ∆Q = Nq where q is the charge on an electron
- I = Nq/∆t
*∆t = length/velocity
*I = Nqv/l
*N (no. of elctrons) = V (volume) * n (no. density) = Al (volume) * n - = Alnqv/l
- = nqvA
Determining resistance of a wire
- Wire connected to ohmmeter
- voltmeter parallel to wire and m ammeter in series, to cell.
Wire resistivity practical
- measure resistance of wire using an ohmmeter for different lengths
- measure diameter of wire using micrometer, & average
- Calculate the cross sectional area
- Plot a graph of resistance against length, compare to equation
- resistivity = gradient * area
Filament bulb IV relationship
*As I increases electrons collide more frequently with lattice resistor ions, transferring more energy/second OR as P.D increases, electrons gain & transfer energy to resistor ions
*Resistor ions to oscillate with greater amplitudes so the temperature of the resistor increases
*Frequent collisions between electrons & lattice ions, limiting drift velocity
*Current, I=nqvA increases by a smaller factor than the P.D (as n, q, A all fixed)
*Resistance is the ratio of pd to current, as pd increases by a greater factor than current does, the ratio R=V/I increases
Temp. increase on thermistor resistance
- Temp. increase electrons gain energy and
- electrons are promoted to conductor
- n increases
- I increases by a greater factor than P.D
- As R = ratio resistance decreases, as current increases by a greater factor than P.D
Light increase on LDR resistance
- electrons gain photon energy from light intensity increase
- electrons are promoted into the conduction band, so n increase
- Current increases
- So, as resistance is the ratio of pd to current, if pd remains constant as current increases, resistance = V/I decreases
Diode IV graph description
- Diode
- Allows current to only flow in one direction
- diode conducts above a threshold voltage
- resistance is high for negative P.D’s, but will conduct for very high -P.D’s
