Electricity Topics: Fields, Resistance, Series and Parallel
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Q1(a)
Define Electric current and quote its units as defined by SI quantities
Q1(b)
In a device if N = 5.4 x 10^6 electrons pass in a time of t = 40 ms determine the current that this equates to
Q2(a)
A calculator uses 50 uA of current in 300 ms with an applied voltage from a battery of 3.5V Calculate the number of electron charge carriers in this time
Q2(b)
What is the potential energy U of the carriers as described in 2(a)
Q3.
In a device the current at any time t can be described by the expression I(t) = I0 exp{-kt}. If at time t = 0 the current is I0 = 8 A and after 14s this drops to 7.2A determine a specific equation describing this behaviour for current and hence determine the total number of charge carriers (electrons) that pass through the device if the device is left on for an infinite amount of time
4(a)
A 15cm x 15cm x 15cm block of pure Silver passes and electric current, determine the resistance of this block in the direction of greatest extent
4(b)
If the block is replaced by a block of Aluminium of the same length of the Silver block determine the dimensions of the end faces so that the block will give an equivalent resistance as determined in 4(a)
5(a)
A wire of diameters 1.8mm and length of 8m passes N = 9 x 10^21 electrons in 1.4 minutes, determine the current density J in the wire
5(b)
Extending from 5(a) determine the potential difference across the ends of the wire
5(c)
Indeed extending on from 5(b) determine the Electric Field (E) in the wire
6(a)
An electrical appliance is rated at a power of 1800 W and uses a rms. voltage of 120V, determine the resistance offered by the appliance
6(b)
From 6(a) determine the current (I) that flows
6(c)
If the appliance is left on for 5 days continuous and the cost of electricity is 5c per KWh determine the cost of leaving the appliance on for this time
7(a)
A circuit consists of 3 pure resistances of 30 Ohms, 60 Ohms and 80 Ohms in series with a 120 V DC supply. Show a circuit representation for this arrangement
7(b)
What is the total resistance of the circuit described in 7(a)?
7(c)
What is the resistance over each of the resistances specified for the circuit in 8(a)?
7(d)
What would be the power dissipated by the series circuit described in 7(a)?
8(a)
A circuit consists of 3 pure resistances of 30 Ohms, 60 Ohms and 80 Ohms in parallel with a 120 V DC supply. Show a circuit representation for this arrangement
8(b)
What is the total resistance R(tot) of the circuit described in 8(a)?
8(c)
Using the total resistance offered by the parallel circuit as determined in 8(b) determine the total current (I)tot drawn from the supply
8(d)
Determine each of the currents that flow through the individual resistor branches, ie. i1, i2 and i3. Hence show that the total current consists of the sum of the branch currents for the parallel network
8(e)
What would be the power dissipated by the circuit of 8(a)?
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Solution Summary
A number of example questions typical of high school and years 1 to 2 electrical science are presented and worked example solutions provided.
The first set of questions involve using parameters such as the number of charge carriers to determine total charge and current through devices and how these formulae can be manipulated to derive such answers as number of charge carriers, total charge, current flowing and current density. The questions and solutions then go on to develop answers for potential difference that occurs across wires and electrical fields that exist as a result.
A number of questions and solutions deal with finding out the Resistance of a material based on resistivity formulae given specific dimensions.
Finally the last set of questions deal with simple resistive series and parallel circuits, and the associated voltages & currents and power dissipated by such circuits.
Solution Preview
Q1(a)
Define Electric current and quote its units as defined by SI quantities
Electric current is defined as the rate of change of electrical charge per second, such that if Q is the electrical charge, t time ( in seconds) then the current is given by dQ/dt. The SI units that define the Ampere are thus quoted in Coulombs per second.
Q1(b)
In a device if N = 5.4 x 10^6 electrons pass in a time of t = 40 ms determine the current that this equates to
N = 5.4 x 10^6 electrons, Time of passage t = 40 ms = 4 x 10^-2 s
One electron carries q = 1.602 x 10^-19 C of charge
Total charge passed Q = N*q = 5.4 x 10^6 x 1.602 x 10^-19
Current is I = Charge/Time = Q/t
Current I = 5.4 x 10^6 x 1.602 x 10^-19 / 4 x 10^-2 A
Current I = 2.16 x 10^-11 A = 21.6 pA
Q2(a)
A calculator uses 50 uA of current in 300 ms with an applied voltage from a battery of 3.5V Calculate the number of electron charge carriers in this time
Current I = 50 uA = 5 x 10^-5 A
Time t = 300 ms = 0.3 s
Charge in time t, Q = I*t = 0.3 x 5 x 10^-5 = 1.5 x 10^-5 C
An electron carries q = 1.602 x 10^-19 C of charge
Number of electron let be N
So
N*q = Q
N = Q/q = 1.5 x 10^-5 / 1.602 x 10^-19
N = 9.36 x 10^13 electrons are passing
Q2(b)
What is the potential energy U of the carriers as described in 2(a)
Electrical potential energy U is given as
U = Voltage x Charge accelerated
U = V*Q
We are told V = 3.5V, We have calculated Q to be Q = 1.5 x 10^-5 C so we can determine U as
U = 3.5 x 1.5 x 10^-5 = 5.25 x 10^-5 J = 52.5 uJ
Q3.
In a device the current at any time t can be described by the expression I(t) = I0 exp{-kt}. If at time t = 0 the current is I0 = 8 A and after 14s this drops to 7.2A determine a specific equation describing this behaviour for current and hence determine the total number of charge carriers (electrons) that pass through the device if the device is left on for an infinite amount of time
I(t) = I0 exp{-kt}
I0 = 8 A the current at t = 0
Thus the current is given by on substitution of I0 = 8 A
I(t) = 8*exp{-kt}
We are told that after 14s, t = 14s, the current I(14) = 7.2
Thus
7.2 = 8*exp{-14k}
exp{-14k} = 7.2/8 = 0.9
Taking log to base e of both sides, ie Natural logs, denoted ln we get
ln(exp{-14k}) = ln(0.9)
Now use the relation ln(exp(x)) = x
Thus
-14k = ln(0.9)
k = - ln(0.9)/14 = 0.00753
Thus we have I0 = 8, k = 0.00753 which we can substitute back into the general equation for I(t) to get
Specific equation for current in time t becomes
I(t) = 8*exp{-0.00753*t}
Now we can integrate this specific expression to get the charge Q
Charge Q = Integral{I}.dt
Q ...
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