Module 4: C10 - Electrical Circuits Flashcards
What is Kirchhoff’s Second Law
In any circuit, the sum of the electromotive forces is equal to the sum of the potential differences around a closed loop.
(Where energy is conserved)
Σε = ΣV in a closed loop.
State Kirchhoff’s First Law and the quantity conserved
Kirchhoff’s first law states that the sum of the current entering a junction is equal to the sum of a current exiting that junction. The physical quantity conserved is the charge. (Conservation of Charge)
State Kirchhoff’s Second Law and the quantity conserved
Kirchhoff’s second law states that the sum of the electromotive forces are equal to the sum of the potential differences around a closed loop. The physical quantity conserved is energy (Conservation of Energy)
When resistors are connected in series, how do you calculate the total resistance.
RT = R1 + R2 + R3
When resistors are connected in parallel, how do you calculate the total resistance.
1/RT = 1/R1 + 1/R2 + 1/R3
Example Question:
Resistors of 6Ω, 4Ω and 3Ω are connected in parallel. Calculate the total resistance of this combination.
Step 1:
Identify the correct equation to calculate the resistance for resistors connected in parallel.
1/RT = 1/R1 + 1/R2 + 1/R3
Step 2: Substitute in known values and calculates a value for 1/R.
1/R = 1/6.0 + 1/4.0 + 1/3.0 = 0.75
Step 3: Invert 1/R to calculate a value for R.
1/R = 0.75 therefore R = 1/0.75
R = 1.3 Ω
Two resistors of resistance 6.9 kΩ are placed in series across a 240V power supply. What is the current flowing through each resistor?
R = 6900Ω
V = 240V
RT = R1 + R2
6900+6900 = 13800Ω
V=IR
I=V/R
240/13800 = 2/115A
= 0.017A
Why would lots of AA Duracell batteries not be able to replace a car battery?
Car batteries have a very low internal resistance so that they can provide the large current needed (often hundreds of amperes) to turn the starter motor in the car. Even if you connected enough AA batteries together to give an e.m.f of 12.6V , the same as a car battery, they would not provide the necessary current because of their internal resistance.
What is Internal Resistance
The amount of energy available to the rest of the circuit (terminal p.d.) is the e.m.f supplied to the cell, minus the energy lost (Lost volts) in the cell as heat.
That lost energy is caused by internal resistance
What does Electromotive Force equal with Kirchhoff’s second law
Electromotive Force = Terminal p.d + lost volts
Example Question:
A typical hand-held torch runs off two 1.5 V cells, yet has a lamp rated at 2.5 V, 0.5 A. Explain how the potential difference across the lamp can actually be 2.5 V as rated. What is the internal resistance of each cell, supposing them to be identical?
1.5 x 2 = 3V
ε = V + Ir
Ir = ε - V
3 - 2.5 = 0.5V
Ir + IR
ε = Ir + IR —> V = ε - Ir
0.5 x r = 3-2.5
0.5 x r = 0.5 —> 1 Ω
Total = r+r —> 1/2 = 0.5 Ω
Example Question:
Calculate the Emf when R1 and R2 are both 4Ω, the internal resistance is 1Ω, and the voltmeter reads 10V.
EMF = ε
R1 + R2 = 8
4 + 4 = 8
Ir = 1
V =10
V = IR
10 = 8R
1.25 Ω = R
ε = V + Ir
ε = 10 + 1.25 = 11.25V
Worked Example:
Calculate the p.d. lost over the internal resistance when Emf =12V, R1 = 4Ω, R2 = 6Ω and R3 = 2Ω.
The reading on the Ammeter says 3A.
(It is a parallel circuit, where R1 is on one line, and R2 and R3 are on another)
1/RT = 1/R1 + 1/R2
1/4 + 1/8 = 3/8
RT = 8/3 = 2.67 Ω
V = IR
V = 3 x 2.67
V = 8
ε = V + Ir
12 = 8 + Ir
4 = Ir
4V = Ir
4V lost over internal resistance
Calculate the voltages over R2 and R3 when Emf = 10V, r = 2Ω, R1 = 5Ω, R2 = 6Ω and R3 = 9Ω.
(It is a parallel circuit, where R1 is on one line, and R2 and R3 are on another)
1/RT = 1/R1 + 1/R2+R3
1/RT = 1/5 + 1/15
1/RT = 4/15
RT = 15/4
RT = 3.75
ε = V+Ir
ε = IR + Ir
10 = I (3.75) + 2
10/5.75 = I
1.74A = I
V = IR
= 1.74 X 3.75
= 6.525V
V2 = 6.525 x 6/15 = 2.61V
V3 = 6.525 x 9/15 = 3.915V
What are Lost Volts and how are they lost
Whenever there is a current in a power source, work has to be done by the charges as they move through the power source.
As a result, some energy is ‘lost’ when there is a current in the power source, and not all the energy transferred to the charge is available for the circuit. The p.d measured at the terminals of the power source is less than the actual e.m.f. We call this difference lost volts.
What is a Potential Divider Circuit
A potential divider is a simple circuit that uses resistors (or thermistors / LDR’s) to supply a variable output potential difference.
Equation for finding the Vout in a Potential Divider Circuit
Vout = (R2 / R1+R2) x Vin
A potential divider is assembled with two fixed resistors, where R1= 1kΩ, R2 = 3kΩ and Vin = 30V. What is the value of Vout?
(When Vout is across R2)
Vout = (3000 / 3000+1000) x 30
Vout = 22.5V
A 270Ω resistor and a 170Ω resistor are connected as part of potential divider circuit to a 36V supply. The output is connected across the 270Ω resistor. Calculate Vout .
Vout = [270 / (270+170)] x 36
Vout = 22.1V
What is ‘Loading a Potential Divider’ (and what can happen as a result of it)
Loading refers to connecting a component or circuit to Vout , that is, placing a component in parallel with R2 . This lowers the resistance of this part of circuit and so lowers Vout.
Note: Adding a large load has little effect on the Vout, but if the load has a small resistance Vout is significantly reduced.
Example Question:
An LDR and a resistor with a resistance of 1000Ω are connected to a 9.0V battery with negligible internal resistance as part of a potential divider. The resistance of the LDR varies from 500Ω in bright light to 50MΩ in darkness. Calculate the minimum and maximum values for Vout.
Minimum:
[500/(1000+500)] x 9 = 3V
Maximum:
[50x10^6/(50x10^6 + 1000)] x9 = 8.99V
What is a Potentiometer (and how does it work)
A potentiometer works in the same way as a potential divider but rather than using two resistors in series it uses a length of resistance wire with a sliding contact.
What happens when the contact is moved upwards (in the direction of A) and downwards (in the direction of B) on a potentiometer.
When the contact is moved towards A, Vout increases, until at A it is equal to Vin. When the contact is moved towards B, Vout decreases until at B it is zero.
What things do you need to remember/reference when approaching how to Solve a Complex Circuit Problem
When tackling any complex circuit problems, we should start with kirchhoff’s laws and these 4 key electrical relationships:
I = Q/t
V = W/Q
P = VI
V = IR
