A 12V, 25W tungsten filament bulb is supplied with current from n cells connected in series. Each cell has an emf of 1.5V and internal resistance of 0.25. What is the value of n in order that the bulb runs at its rated power?
Take a look at the titles of all the other threads. You can see that a title of help! is of no help to anybody.
Hello, Just "Help !" as title is not very inspiring. A more descriptive title will draw more attention. Have a look at this page about batteries and their internal resistances: http://www.allaboutcircuits.com/vol_1/chpt_11/3.html Bertus
Hello, Your picture is rather fuzzy and not good to read. When you must make a battery of 12 Volts, out of 1.5 Volt cells. How many cells are needed? The internal resistance of the cells can be added. Bertus
Hello, Did you read this before posing in the homework section? PLEASE READ - Posting Questions in the Homework Help Forum Show us any effort and we will help you. I already gave you a hint about the battery. Bertus
I've tried it again and again but iam not getting anywhere near answer. You said add the internal resistance, but then again there is another question how many times do I add the internal resistance? Or can I just simply divide 12V/1.5V?
Hello, For the number of cells you can indeed divide the 12 Volts by the 1.5 Volts. Then you can also calculate the internal resistance. Bertus
Well that is not entirely true, if the resistance of each battery was for example 0,33 ohm, then the output power would be closest to 25w with 15 cells instead of 12, so you need to take that into account.
Lets say you have 8 batteries in series, that gives you 12V. But since each battery has 0.25 ohms, the total resistance inside the battery will be 0.25*8=2 ohms. If you draw 0.5A from that battery you will have only 12-2*0.5=11V volts available at the battery terminals. 25W at 12V is a bit over 2A, so you can imagine it will take more than 8 batteries to get the output voltage near 12V when providing 2A of current. The big question is how many cells do you actually need to get that output voltage under load.
First you need to find what the current is when you have n cells. Then use P=R*I^2 to find out the power going into the bulb.
I get what you are saying but in order to find the current, shouldn't I know the value of n? Or the current would remain same throughout. Can I do it like this. P=V^2/R 25=12^2/R R=5.76ohm Now V=IR 12/5.76=I I=2.08amp Is this the value of current?
You need to start from the other side. The voltage will be n*1.5V and the internal resistance will be n*0.25ohms. To this voltage you have connected the internal resistance and load resistance in series, what will the current be? This current is then used to calculate the power across the load resistor, you should get an equation with n as the only variable, then you solve for n that gives 25W. Then find the two adhacent whole numbers and pick the one that gives closer power.
Q1) How much current does the bulb require? Q2) If a single cell is delivering the current in Q1, what will the voltage of that cell be? Q3) What is the smallest number of cells, each at the voltage in Q2, that can supply at least 12V to the load.