Battery modelling

6673 /www/nrich/html/content/id/6673/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/7456">Warm-up</a></li> <li><a href="http://nrich.maths.org/8690">Try this next</a></li> <li><a href="http://nrich.maths.org/6629">Think higher</a></li> <li><a href="http://nrich.maths.org/6661">Read: mathematics</a></li> <li><a href="http://en.wikipedia.org/wiki/Internal_resistance">Read: science</a></li> <li><a href="http://en.wikipedia.org/wiki/Power_engineering">Explore further</a></li> </ul> <div> </div> A battery is modelled as a 24V EMF in series with an internal resistance of $2\Omega$. The battery is being charged by a constant current source as shown in the network below. The ammeter, used to measure the current through the battery, has a voltage drop of 2V at full range scales. Here it is set to a 20A range and reads a current of 2A. What is the value of the constant current supplied? What fraction of the input power is supplied to the battery? What current would flow into the battery if we now replaced the ammeter with a wire of zero impedance? <mdo:image alt="" height="537" src="Battery.jpg" width="861"></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <mdo:image height="488" width="525" src="Battery1.jpg" alt=""></mdo:image> Part 1: The ammeter will drop a voltage of 2V at 20A, it can therefore be modelled as a resistor of 0.1$\Omega$. The $100\Omega$ resistor is in parallel with the series combination of the ammeter and battery. $100 I_1 = (0.1x2) + (2x2) + 24$ $I_1 = 0.282 A$ Current Conservation: $I_0 = I_1 + I_2$ $I_0 = 2 + 0.282 = 2.282 A$ Part 2: Power supplied = Power consumed Power Supplied = $ (0.282^2 x100) + (2^2 x 0.1) + (2x2^2) + (2x24)$ = 64.35 W Power dissipated by battery = IV = 2 x 24 = 48W 74.6 % of the power supllied is fed into the battery. Part 3: $I_0$ is unchanged, $I_0$ = 2.282A Current conservation: $I_0 = I_1 + I_2$ Voltage Conservation: $100 I_1 = 24 + 2I_2 $ Solving these two equations simultaneously we find $I_2 = \frac{I_0 - 0.24}{1.02} = 2.002 A$ </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Kirchoff's current law states that the sum of the currents into a node are equal to the currents exiting. Kirchoff's voltage law states that the sum of the voltages around a closed loop equals zero. </mdoxml> 2 3 0 0 0 0 1 Battery modelling Find out how to model a battery mathematically Applications engineering Applications physics Algebra Simultaneous equations Applications engineering Admin Individual