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Aim: To determine the electromotive force (EMF) and internal resistance of a standard dry cells. Introduction: Electromotive force is the opposite of potential difference where a voltage is gaining energy. This seems unusual, but it is required in order to allow an electric circuit to function. An electric current is a flow of electric charge, the charge flows around the circuit, transferring some of its energy to region of resistance along the components (resistors, bulbs, buzzers). At some point along the circuit the energy must be initially supplied, or else the whole process could not function. At the beginning of the process the power supply provides the charge with energy to pass to the circuit which results electromotive force. EMF is a type of voltage along with potential difference and these are defined as ;Voltage where the charge is losing energy is a potential difference, V. A voltage where the charge is gaining energy is an electromotive force, E. A device which can maintain different levels of electrical potential (voltage) across its terminals is said to be a source of electromotive force or emf, E. In Diagram 1 below, terminals a and b are connected to an external circuit with a resistor R. When the emf source is turned on, positive charge is driven around the circuit from point a to point b through the resistor R. Real voltage sources such as batteries, have an internal resistance (r) which causes the external terminal potential difference, V ab, to be less than the emf supplied by the source.

Diagram 1

Apparatus * Dry cells - Standard Laboratory 12v * Leads - Copper insulated wires * Ammeter - 0-10A * Voltmeter - 0-10V * Rheostat - 5 amps, 20 Ohms max. Circuit Analysis: 1. Use Ohmâ€™s Law (Vtotal=IRtotal) and find an expression relating the EMF to measure current and total resistance 2. Solve the expression for IR 3. E.M.F = (V - I ) / r Method: 1

Arrange apparatus as shown in the below circuit diagram

2

Start off by recording the corresponding voltage (terminal pd) and current figuring

out the values available in the voltmeter and ammeter. 3

Repeat previous procedures while the external load is varied, so that you obtain a set

of voltage and current readings. 4

Calculate averages and plot a graph of current against voltage. Use the graph to

figure out the internal resistance and the EMF of the power supply.

Data Collection Terminal Voltage (V) 1.09 1.65 1.84 1.84 2.06 2.21 2.22 2.28 2.33 2.39 2.40 2.47 2.49 2.52 2.59 2.63 2.75 Data Presentation

Current (I) 1.38 1.04 0.83 0.81 0.75 0.63 0.59 0.56 0.54 0.50 0.46 0.40 0.37 0.35 0.31 0.26 0.11

Conclusion Looking at my results, I found that the voltage of the circuit was increased as the current reduced. This, as according to my graphs, leads to a increase in resistance from the slope of the graph which is the gradient. My predication, based on background research and previous knowledge, was accurate and indicated the results. This increment follows ohms law, as the increases is proportional to the other calculated results. From the graphs I can calculate both the e.m.f and internal resistance for the power supply. The y-intercept gives me the value for I, and the gradient of the slope gives me the value for -r, a positive value of the internal resistance, using this method gives a fairly accurate result even though the result was limited by the voltmeter's two decimal place display. I also noticed that the internal resistance changed when using higher voltage reading on the dry cells, this was due to an increase in current. Thus, I have noticed that r cannot be simply described as an extra resistor located inside the power supply. The internal resistance is small in comparison to the external resistor, this allows for an efficient energy transfer. Evaluation I have noted that the digital meters gave good result of the rheostat in order to control the amount of the current. The dry cell kept the voltage in the circuit at a stationary level before it is changed by the alteration of the current, these features combined meant that the procedures taken during this experiment were suitable for this investigation. The investigation showed no doubt results, which can be attributed to the repetition of results in order to determine an average. This as evidence that the data is consistent .There was also an issue when trying to take reading at low temperatures due to the nature of rheostat sensitivity which resulting in a lower voltage range and cannot be shown on graph.Nevertheless, it can be improvised by using a rheostat with a greater range of selections, to reduce the error on lower voltages. There were many limitations that I encountered during the conduction of this experiment; mainly this involved the equipment and digital readings. Replacing these with significantly more accurate pieces

of equipment would reduce the error range for the gradient, therefore giving more accurate final results.

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Diagram 1

Apparatus * Dry cells - Standard Laboratory 12v * Leads - Copper insulated wires * Ammeter - 0-10A * Voltmeter - 0-10V * Rheostat - 5 amps, 20 Ohms max. Circuit Analysis: 1. Use Ohmâ€™s Law (Vtotal=IRtotal) and find an expression relating the EMF to measure current and total resistance 2. Solve the expression for IR 3. E.M.F = (V - I ) / r Method: 1

Arrange apparatus as shown in the below circuit diagram

2

Start off by recording the corresponding voltage (terminal pd) and current figuring

out the values available in the voltmeter and ammeter. 3

Repeat previous procedures while the external load is varied, so that you obtain a set

of voltage and current readings. 4

Calculate averages and plot a graph of current against voltage. Use the graph to

figure out the internal resistance and the EMF of the power supply.

Data Collection Terminal Voltage (V) 1.09 1.65 1.84 1.84 2.06 2.21 2.22 2.28 2.33 2.39 2.40 2.47 2.49 2.52 2.59 2.63 2.75 Data Presentation

Current (I) 1.38 1.04 0.83 0.81 0.75 0.63 0.59 0.56 0.54 0.50 0.46 0.40 0.37 0.35 0.31 0.26 0.11

Conclusion Looking at my results, I found that the voltage of the circuit was increased as the current reduced. This, as according to my graphs, leads to a increase in resistance from the slope of the graph which is the gradient. My predication, based on background research and previous knowledge, was accurate and indicated the results. This increment follows ohms law, as the increases is proportional to the other calculated results. From the graphs I can calculate both the e.m.f and internal resistance for the power supply. The y-intercept gives me the value for I, and the gradient of the slope gives me the value for -r, a positive value of the internal resistance, using this method gives a fairly accurate result even though the result was limited by the voltmeter's two decimal place display. I also noticed that the internal resistance changed when using higher voltage reading on the dry cells, this was due to an increase in current. Thus, I have noticed that r cannot be simply described as an extra resistor located inside the power supply. The internal resistance is small in comparison to the external resistor, this allows for an efficient energy transfer. Evaluation I have noted that the digital meters gave good result of the rheostat in order to control the amount of the current. The dry cell kept the voltage in the circuit at a stationary level before it is changed by the alteration of the current, these features combined meant that the procedures taken during this experiment were suitable for this investigation. The investigation showed no doubt results, which can be attributed to the repetition of results in order to determine an average. This as evidence that the data is consistent .There was also an issue when trying to take reading at low temperatures due to the nature of rheostat sensitivity which resulting in a lower voltage range and cannot be shown on graph.Nevertheless, it can be improvised by using a rheostat with a greater range of selections, to reduce the error on lower voltages. There were many limitations that I encountered during the conduction of this experiment; mainly this involved the equipment and digital readings. Replacing these with significantly more accurate pieces

of equipment would reduce the error range for the gradient, therefore giving more accurate final results.

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