Answer Happy

THE EMF AND INTERNAL RESISTANCE OF A SOURC 1/3 Aims The aims of the experiment are to determine: a) the emf, E, and internal resistance, r, of a cell; b) the condition for maximum power transfer. Method The small box provided contains a cell of emf E and internal resistance r. Using this cell, an ammeter (with negligible resistance), a variable (but known) resistance (R) and a plug key connect the circuit as shown below. Note the meter polarity. min to Does it matter in what order the components are connected? Explain. In this experiment I is the dependent and R is the independent variable. Is this correct? Explain. Write down the equation for E in terms of the current, 1, the external resistance, R, and the internal resistance of the cell r E2.1 Show that equation (E2.1) can be rearranged into a form suitable for plotting a straight-line granh: 1 - 1 R+ E E E2.2 1

- Explain what should be plotted against what in order to get a straight-line graph, and what you can determine from the slope and the intercepts on both axes. Is it necessary to show the origin on the graph if both intercepts are to be read off? Explain. Take current readings for values of R ranging from 20 92 to 100 12 in steps of 20 12, and then up to 400 12 in steps of 50 2. (Do not have R< 20 12, as this would draw too much current from the battery) Tabulate your readings and the quantities that you calculate from them in the table below. You should convert the milliammeter readings to amperes. Included also is a column for the power dissipated in the external resistance P=1°R. R (12) I (mA) * (A+) P=ľR(W) 20 75.3 40 66.0 60 60.3 80 53.4 100 48.8 150 41.5 200 35.2 250 28.2 300 20.0 350 13.1 400 0.0

Analysis (a) Plot the straight-line graph suggested by equation (E2.2). N.B. Make sure that you can read a sufficiently large negative intercept on the R axis. From the graph, showing your calculations, determine & and r (the latter in two ways). (b) Plot P against R. Determine the value of R for which P is a maximum. Compare this value with that of r. What do you conclude from this comparison? - Explain whether or not this graph should go through the origin.

Experiment / Measurements: Take readings of battery terminal voltage (V) for each step increment of current (1) with 0.6 (A). Adjust the resistance of your resistor box accordingly for current 0.6 (A) increment stepwise. V 1(A) V (V) Cell or Battery 0 6 & 0.6 5.76 1.2 5.52 A 1.8 5.28 R 2.4 5.04 Circuit Diagram

Analysis E The terminal voltage is equal to IR so Eq.1 can be rearranged as: V = - Ir + ε → like y = mx + c gradient or -OOOOOOO = -r When a graph plotted V against, I should give you straight line with gradient - r. . The intercept on the y-axis is equal to the emf of the cell. The gradient or slope of the plot is equal to -r and r is the internal resistance of cell I/A . emf is very important for electrical generators such as hydro dams, transformers used to boost the voltage.

Objectives 1. Find emf (?) and internal resistance (r) of a cell/battery (energy source) in a closed circuit. Ohm's Law: V = IR Electromotive Force emf is the energy provided by a cell or battery, per coulomb of charge passing through it and it is measured in volts (V). Kirchhoff's 2nd Law: As charge goes around the circuit the sum of EMFs must equal the sum of voltage drops leading to: E = 1 (R + r) = IR + Ir = V + Ir → E.q 1 Internal Resistance Batteries or Cells have an internal resistance (r) measured in ohm's. When electricity flows around a circuit the internal resistance of the cell itself resists the flow of current and so thermal energy is wasted in the cell itself. = ({ – V)/1- → E.q2 r OOOOOO