# Investigating the E.m.f and Internal Resistance of 2 cells on different circuit Structures.

Khaled Hamid Page

Investigating the E.m.f and Internal Resistance of 2 cells on different circuit Structures

Background Knowledge:

For this investigation the main variable will be the circuit structure and positioning of the two cells. Therefore I will discuss in brief the most relevant aspects of the electricity module regarding this investigation. These are the electrical circuits, the electromotive force and the internal resistance.

Circuits: According to Kirchoff’s first law, the series circuit is one in which the components are connected one after another, forming one complete loop.

Diag1:

The next circuit I will also be using is the parallel circuit. I am using these two circuits because the main aim from this investigation is see how the two circuits influence the e.m.f and the internal resistance.

The parallel circuit is one where the current can take alternative routes in different loops. The current divides at a junction, but the current entering the junction is the same as the current leaving it.

Therefore: I = I1 + I2 + I3

Diag1:

Electromotive force

The second factor involved in the investigation is the electromotive force. When charges pass through a power supply such as a battery, it gains electrical energy. The power supply is said to have an electromotive force (e.m.f). The electromotive force measures in volts, the electrical energy gained by each coulomb of charge that passes through the power supply. E.m.f is not actually a force however. The energy gained by the charge comes from the chemical energy of the battery.

e.m.f = energy converted from other forms to electrical

charge

The electromotive force in a closed circuit is also equal to the sum of the potential difference. Therefore:

E = IR + Ir

R – external resistance r – internal resistance

The number of large R’s depends on the number of external resistors in the circuit

Internal Resistance:

All power supplies have some resistance between their terminals called internal resistance. This causes the charge circulating in the circuit to dissipate some electrical energy in the power supply itself. The power supply becomes warm when it delivers a current.

Diag 3:

The circuit above shows a power supply which has e.m.f (E) and internal resistance (r). It delivers a current I when connected to an external resistor of resistance R, also called the load. Vr is the potential difference across the internal resistance. Kirchoffs second law states:

E = Vr + VR

The potential difference VR across the load is thus given by

VR = E – Vr

VR is called the terminal potential difference.

Returning to diagram 3, V = IR and Vr = Ir, so E = Vr + VR which becomes:

E = IR + Ir or E = I (r + R)

Quite often, the internal resistance is considered negligible but for this investigation, it will be highly considered and part of a constant battery cell.

Lost volts

The greater the current delivered by the power supply, the lower its terminal potential difference. If more components are connected in parallel to the power supply, the current increases and the lost volts are given by:

Lost volts = current * internal resistance

Aim:

The aim of this investigation can be derived from the title. I will conduct an investigation on how the e.m.f and internal resistance are influenced by the changing of the circuit and where the cells are positioned, and I will also use external resistors to clearly identify any influences or alterations

Hypothesis:

For this investigation I have a possible hypothesis which seems plausible using the background knowledge I know.

In a series circuit with two cells, I expect my readings to obey ohms law; when the resistance increases the current decreases. I also expect the e.m.f to be double the e.m.f of one cell in series. I decided this prediction from the equation of e.m.f = Ir + Ir, therefore for two cells the e.m.f should double and I got the equation (when external resistor = 12 ohm)

e.m.f = (I*2r) + potential difference of external resistor

- using figures e.m.f = (2*1*0.5) + (0.5*12)

E = (1 + 6) = 7 V

I also expect the internal resistance to be double the internal resistance of a parallel circuit with two cells because in a parallel circuit the resistors are halved so therefore in series the internal resistance is (r). If r was 2 ohms:

In parallel the internal resistance is (1/r + 1/r). If r was 2ohms again then internal resistance would be 1ohms. This is half of the internal resistance of a series circuit with two cells.

Diag series with two cells:

The series with one cell will have half the values of the series with two cells, because it is the same circuit but with half of the number of cells so internal resistance and e.m.f should be exactly half of the series circuit with two cells

e.m.f = Ir + IR = (0.25*12) + (0.5 * 1) = 3.5 V (exactly half of the series value with two cells above)

Diag series with one cell:

In a parallel circuit, I expect the e.m.f to be the same as the e.m.f of the series circuit because the same amount of voltage will be needed to ‘push’ the current around the circuit.

I expect the e.m.f to be; (using figures where external resistor is 12V)

e.m.f = (2* p.d of internal resistor) + (p.d of external resistor)

E = (2* 1* 0.5) + (12 * 0.5)

E = 7V

(I am little sceptical about this prediction but hopefully I will prove myself wrong/right later in the investigation)

I expect the internal resistance to be halved in the parallel circuit, because in series circuit the internal resistance is r = r1 + r2. However in parallel, the internal resistance is:

1/r = 1/r1 + 1/r2 = r = r1 + r2/ 2