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I already asked sagarhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf
no 36 (D) ,no 38 (C) thank youu!!
36)
Any two conductors connected to each other by another conductor will have the same potential.
This is like having two separate tubes of water (each one represents one conductor), with different amounts and different heights in each (the height represents the potential, the amount represents the charge) - if they are connected by a tube or any other medium that allows free flow of water (this medium represents the conductor) then the heights will become equal (the potential will become equal).
Note that this does not mean "potential" flows from one to another - just like "height" does not flow, potential won't. The charges on an object make up it's electric potential, and these charges will flow (since all materials involved are conductors).
In the analogy, water (the charges) flows to equalize the levels.
In the end, the volume of water in each tube may be different (i.e. the charge on each conductor may be different) but the height (potential) will be the same.
However, this potential only becomes equal in a static situation.
So we can say the potential at point X is +24 Volts because it is connected to the positive terminal of the source, which is at a potential of +24 Volts itself.
Suppose current is flowing through the motor, energy is lost in the motor until the potential of the current becomes equal to the potential of the negative terminal. So the potential at Y is 0 Volts.
The problem arises when either wire is cut - suppose you cut the positive cable somewhere along it's length. What is the potential of X then? It is not connected to the 24 Volt terminal, so what else could the potential there be?
The answer is that the potential of X is 0 Volts, since it is connected to the negative terminal of the battery, which we have assigned to have a zero potential. When the wire is cut no current can flow through the circuit, so the resistance inside the motor stops acting like a resistance and acts like a normal conductor, which means the point X will be at 0 Volts.
Suppose we cut the negative cable. Following the same logic, since the motor no longer does anything (no current because circuit is incomplete), it simply acts as a conductor and equalizes the potential at X and Y, so that both have a potential of +24 Volts.
Suppose the connection in the motor breaks, again no current will flow through the circuit, and X will be at a potential of +24 Volts (since it is still connected by an unbroken connector to the positive terminal of the source) with Y remaining at 0 Volts (since it is connected to the negative terminal of the source).
So having seen all this, we can say that D is the only right answer among all the options.
Let me know if you have any doubts, since it was a little difficult to explain this one, and the concept is not so straightforward anyways.
38)
This is a neat little question, with the only law needed being Kirchoff's Second Law.
There is a very, very quick way of solving it, but first the long way:
i) Suppose you start at the bottom right corner of the circuit.
ii) You go left, across the battery, and find an increase in potential of 20 Volts.
iii) Okay, now you go up (no resistance), go right till the junction (again, no resistance) and take the upper branch.
iv) As you pass the resistor L, you see a drop of 7 Volts.
v) You continue across M where there is some unknown change in potential and then
vi) return to the bottom right corner.
So, the total change in potential should be zero. Therefore,
+20 Volts + (-7)Volts + (Change in potential across M) = 0
13 + (change in potential across M) = 0
Therefore, the change in potential across M = -13 Volts. The change is -13 Volts, the drop is 13 Volts. So B or C.
Let's narrow it down using Q.
i) Again, start at the bottom right corner of the circuit.
ii) Go left, cross the battery, and see an increase in 20 Volts.
iii) Continue until you reach the junction, and pick the upper branch.
iv) Cross P, seeing a drop of 7 Volts, and take the bridge between the branches.
v) Cross N, seeing a drop of 4 Volts, and go towards Q, i.e. to the right on the diagram..
vi) There is some unknown change in potential across Q, and
vii)you then return to the bottom right corner.
Again, the total change in potential should be zero. So,
+20 Volts + (-7) Volts + (-4) Volts + (Change in potential across Q) = 0
9 Volts + (Change in potential across Q) = 0
Change in potential across Q = -9 Volts. Therefore drop = 9 Volts, so the only option that has that is C.
The thing is, that you could have gotten the answer directly without the first part if you had just gone past Q - you would get 9 Volts, and the only option with a 9 Volts drop for Q would be C!
But be careful here - all these changes are so simple because we are going in the direction of current. Suppose you go across a resistor with this method and you are going in the direction opposite to the current, then you will have anincrease in potential, not a decrease.
This is because current flows from a region of high potential to low potential - suppose you are going along the current, you are also going from the region of high potential to the region of lower potential. But if you go in the direction opposite to the current you are going from a region of low potential to a region of high potential, which is an increase in potential - you have to make sure you adjust the signs correctly.
