Open this link to answer fig22.17 If you are not sure (or even if you think you
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fig22.17
If you are not sure (or even if you think you are sure) about the above question, open the Circuit Simulator on Moodle and build the above circuit. You can "measure" the currents at A and B simply by counting the charges (the little blue balls) that pass by A and B over some set amount of time (say 10 s). Or you can look at how the charges behave and deduce the answer. In Circuit Simulator on Moodle. Play around until you are comfortable (in particular, note that right clicking on a junction between components gives you the option of disconnecting the junction). Now set up the circuit in the diagram at right. When the switch is "open" (no charge flowing through it) compare the brightnesses of bulbs A and C. What happens to the brightness of bulb C when you close the switch? When the switch is closed (current is able to flow through it) look at what is happening at the junction where the two branches to lightbulbs B and C split. Explain how the conservation of current is being depicted by the simulation. Clear the circuit that you have built (just press the "Clear" button). Build a circuit in which there is just a battery and a series of wires running around from one battery terminal to the other (you are building a circuit like the ones in Fig. 22.17 in your text). This is a "short circuit". When you first build it you will find that the battery will light on fire (!). This is because the current through the battery is so large that its internal resistance causes it to heat up by dangerous amounts (don't try this with a real battery! A burning battery is VERY dangerous!) In the "Advanced" part of the control panel press the button that says "Show". The wire resistivity will be set to "Almost none". Turn up the wire resistivity until the current is low enough that the battery isn't on fire. Leaving the resistivity constant, how can you change the wires so that the current increases? Do not introduce any new current paths. You are simply modifying some property of the wires that you already have. Hint: I'm asking you to decrease the resistance of the wires without changing the resistivity. Check that your idea works by trying out your idea and seeing if the battery lights on fire again. At right is a figure showing a simple circuit. A potential vs. position plot, similar to the ones in section 22.5, has been drawn next to it. However, there is a serious error in the potential vs. position plot. What is the serious error? The current is determined by the potential difference and the resistance of the wire: I = DeltaVchem/RExplanation / Answer
1. When the switch is open A perticular current will pass through series connection of Bulb A and C which will lead to glow in A and C and B will remain OFF!
2. When the switch is closed B and C will be connected in parallel and hence current through C will decrease leading to decrease in brightness of bulb C
3. As explained earlier on closing the switch the net resistance due to parallel combination of B and C gets reduced and more current passes through A which then gets divided between B and C
So Current through B+ Current through C=Current through A
4.SInce Resistance = resistivity*l/A
where l is length and A is area of cross section So to increase the current resistance needs to be Decreased! TO do so We can Either Increase the area of cross section A of the wire or Decrease the length of wire.
5. THe plot shows two different potential of wire from battery to resistance and wire from resistance to battery. The potential of wire from resistance to battery must reach till 0v since it is connected to the negetive/0 potential of the battery!
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