Module 4: C10 - Electrical Circuits

Question 1

What is Kirchhoff’s Second Law

Answer 1

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.

Question 2

State Kirchhoff’s First Law and the quantity conserved

Answer 2

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)

Question 3

State Kirchhoff’s Second Law and the quantity conserved

Answer 3

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)

Question 4

When resistors are connected in series, how do you calculate the total resistance.

Answer 4

RT = R1 + R2 + R3

Question 5

When resistors are connected in parallel, how do you calculate the total resistance.

Answer 5

1/RT = 1/R1 + 1/R2 + 1/R3

Question 6

Example Question:

Resistors of 6Ω, 4Ω and 3Ω are connected in parallel. Calculate the total resistance of this combination.

Answer 6

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

Question 7

Two resistors of resistance 6.9 kΩ are placed in series across a 240V power supply. What is the current flowing through each resistor?

Answer 7

R = 6900Ω
V = 240V

RT = R1 + R2
6900+6900 = 13800Ω

V=IR
I=V/R

240/13800 = 2/115A

= 0.017A

Question 8

Why would lots of AA Duracell batteries not be able to replace a car battery?

Answer 8

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.

Question 9

What is Internal Resistance

Answer 9

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

Question 10

What does Electromotive Force equal with Kirchhoff’s second law

Answer 10

Electromotive Force = Terminal p.d + lost volts

Question 11

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?

Answer 11

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

Question 12

Example Question:

Calculate the Emf when R​1​ and R​2​ are both 4Ω, the internal resistance is 1Ω, and the voltmeter reads 10V.

Answer 12

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

Question 13

Worked Example:

Calculate the p.d. lost over the internal resistance when Emf =12V, R​1 =​ 4Ω, R​2 =​ 6Ω and R​3 =​ 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)

Answer 13

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

Question 14

Calculate the voltages over R​2​ and R​3​ when Emf = 10V, r = 2Ω, R​1 =​ 5Ω, R​2 =​ 6Ω and R​3 =​ 9Ω.

(It is a parallel circuit, where R1 is on one line, and R2 and R3 are on another)

Answer 14

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

Question 15

What are Lost Volts and how are they lost

Answer 15

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.

Question 16

What is a Potential Divider Circuit

Answer 16

A potential divider is a simple circuit that uses resistors (or thermistors / LDR’s) to supply a variable output potential difference.

Question 17

Equation for finding the Vout in a Potential Divider Circuit

Answer 17

Vout = (R2 / R1+R2) x Vin

Question 18

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)

Answer 18

Vout = (3000 / 3000+1000) x 30

Vout = 22.5V

Question 19

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 .

Answer 19

Vout = [270 / (270+170)] x 36

Vout = 22.1V

Question 20

What is ‘Loading a Potential Divider’ (and what can happen as a result of it)

Answer 20

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.

Question 21

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.

Answer 21

Minimum:
[500/(1000+500)] x 9 = 3V

Maximum:
[50x10^6/(50x10^6 + 1000)] x9 = 8.99V

Question 22

What is a Potentiometer (and how does it work)

Answer 22

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.

Question 23

What happens when the contact is moved upwards (in the direction of A) and downwards (in the direction of B) on a potentiometer.

Answer 23

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.

Question 24

What things do you need to remember/reference when approaching how to Solve a Complex Circuit Problem

Answer 24

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