Module 4 Electrical Circuits

Question 1

How can you represent Kirchhoff’s first law using algebra?

Answer 1

Ii; currents entering a junction
Io; currents leaving a junction.

Sum of all Ii + Sum of all Io (negative values)=0

Question 2

State Kirchhoff’s second law.

What law of conservation is this law a consequence of?

Answer 2

The sum of emf’s are equal to the sum of product of resistance and current across a circuit.
It is a consequence of the law of conservation of energy.

Question 3

While doing calculations using Kirchhoff’s second law, how should you divide a parallel circuit and what precautions do you need to take to ensure a correct answer?

Answer 3

Divide the circuit in series loops and calculate the emf and voltage for each loop.
State one direction of flow of current as positive.

Question 4

How can you mathematically use Kirchhoff’s second law to calculate unknown current?

Answer 4

After you’ve written equations for each loop in a circuit, you can treat these as simultaneous equations and make current the subject.

Question 5

Show how the effective resistance of a series circuit can be found out using Kirchhoff’s second law.

Answer 5

By Kirchhoff’s second law Emf=V across circuit
Rt= Total resistance
As current is constant across series circuits.
I x Rt=(Ir1)+(Ir2)+…
I can cancel out leaving behind,
Rt= r1+r2+…

Question 6

Derive the equation to calculate the total resistance in parallel circuits.

Answer 6

It=I1+I2+...
I=V/R
Vt/Rt=v1/r1+v2/r2+...
Vx(1/Rt)=Vx(1/r1+1/r2+...)
1/Rt=1/r1+1/r2+...

Question 7

What happens to total emf when a 5V battery provides emf in a clockwise direction and a 20V battery provides emf in an anti-clockwise direction?

Answer 7

Current of 15V flows anti-clockwise.

Question 8

What does a potential divider do and how?

Answer 8

It splits the voltage from a source of emf and controls how much voltage is applied through a component.
The v supplied by a power source is split between two loads in series.

Question 9

What equation is used to calculate V out in a potential divider?

Answer 9

Vout= Vin x R2/r1+r2

Question 10

What is a potentiometer?

Answer 10

A potential divider that uses a variable resistor instead of 2 fixed resistors in series.

Question 11

Describe two uses of a potentiometer circuit.

Answer 11

Controlling street lights, ldr used as final resistor in series .
High resistance at night time will require high voltage across R1, giving a higher Vout powering a street light.

Controlling heaters, NTC thermistor used as final resistor in series.
High resistance at low temperatures, pulling more voltage. This increases Vout and turns heater on.

Question 12

Is the ratio of voltages across resistors in a potentiometer equal to the ratio of their resistance?

Answer 12

Yup.

Question 13

What is terminal pd?

How does it relate to emf?

Answer 13

It is the pd across the terminals of a source of emf.

Terminal pd when switch is open is equal to emf.

Question 14

What are lost volts?

How are they related to terminal pd and emf.

Answer 14

This is energy lost due to internal resistance of a power source.
It is equal to the difference between Emf and terminal pd.

Question 15

What is internal resistance and how do you relate it to emf mathematically?

Answer 15

It is the resistance to flow of charge within a power source because of it’s internal structure.
Emf=IR + Ir

Question 16

Describe how you would investigate internal resistance of a cell.

Answer 16

By using equation emf=Ir+IR
VR=IxR, VR is voltage across circuit.
emf=Ir+VR
We can rearrange the equation to
VR=-rI + emf
We can then plot VR on the y axis and I on the x axis. 

We can measure the terminal pd using voltmeter across cell and attach cell to variable resistor. Change the value of resistance and measure changes in V and I using ammeter in series to the circuit.

Find gradient of which is equal to -r