Question 1
A potentiometer circuit that is used
as a means of comparing potential differences is shown in Fig. 5.1.
Fig. 5.1
A cell of e.m.f. E1 and
internal resistance r1 is connected in series with a resistor of resistance R1
and a uniform metal wire of total resistance R2.
A second cell of e.m.f. E2
and internal resistance r2 is connected in series with a sensitive ammeter
and is then connected across the wire at BJ. The connection at J is halfway along
the wire. The current directions are shown on Fig. 5.1.
(a) Use Kirchhoff’s laws to obtain the relation
(i) between the currents I1,
I2 and I3, [1]
(ii) between E1, R1,
R2, r1, I1 and I2 in loop HBJFGH, [1]
(iii) between E1, E2,
r1, r2, R1, R2, I1 and I3
in the loop HBCDJFGH. [2]
(b) The connection at J is moved along the wire. Explain why the
reading on the ammeter changes. [2]
Reference: Past Exam Paper – November 2011 Paper 22 Q5
Solution 1:
(a)
(i) I2 = I1 +
I3 (Kirchoff’s
1st law: Sum of current entering a junction = sum of current leaving
the junction)
(ii)
{Put e.m.f of the cell on
one side. Since a current I1 is coming from the +ve terminal of the
cell, the same current should come back to (negative terminal of) the cell. So,
current I1 flows through r1 and R1 which are
in series.
At junction B, the current
I1 changes into I2 and I3 (according to the
above equation) but since we are considering loop HBJFGH, we can neglect I3.
From B to J, the current I2 flows. Since J is halfway on the wire,
half of the total resistance of the wire (R2) should be considered
[since resistance is directly proportional to the length of wire –> R = ρL / A]. So, the p.d. across BJ = I2(R2/2).
Now, consider from the
junction J to part F, then G up to the negative terminal of cell E1.
There is no more junction where the current changes. So, the current going back
to the back is the same as that flowing through part JF of the wire (that is, I1).
And again, since J is halfway on the wire, part JF would correspond to a
resistance of R2/2. So, the p.d. across JF = I1(R2/2).
Hence, the equation below is obtained.}
E1 = I2(R2/2)
+ I1(R2/2) + I1R1 + I1r1
(iii)
{Tip: Put emf of cells (E1
and E2) on one sides of equation and potential differences in
components on other side. (Kirchoff’s 2nd law)
E2 is in
opposite direction of E1 (check direction of currents I3
and I1).
I2 not present
since we are not considering BJ. Current coming from J towards F is I1
(corresponding to pd = I1R2/2). Current across r2
in JDCB is I3(corresponding to pd = I3r2). In
GH, pd = I1R1 + I1r1.}
E1 – E2 = -I3r2
+ I1(R1 + r1 + R2/2)
(b) The potential difference across BJ of the wire changes /
resistance of BJ changes. There is a difference in potential difference
across the wire and the potential difference across cell E2.
{Since V = IR, current I =
V / R. As there is a difference in V, the current also changes}
this blog is really VERY helpful! thanks a lot. :)
ReplyDeletePls help me....
ReplyDeleteOn a ii..
Where does the I2 vanishes after BJ ???
Shouldn't I2 go with I1 well???
After junction J (to F, G ...), the current becomes I1 again (as explained above). This again satisfies the equation in a (i).
DeleteI really don't understand why in 3rd part we are subtracting the emfs. How are they both in opposite direction? (For that shouldn't positive terminal be connected to negative terminal?)
ReplyDeleteHow are the emfs in opposite direction. To subtract them shouldn't the positive terminal be connected to positive.
ReplyDeleteI'm really confused in this.
The correct way to connect cells is to connect the positive terminal of one cell to the negative terminal of the other. If the +ve terminal of one is connected to the negative terminal of the other, the connection is incorrect. We then need to subtract the e.m.f.s (bigger emf – small emf).
DeleteIn the circuit, the positive terminals of both cells are connected at junction B. So, we need to subtract the emf.