__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. E

_{1}and internal resistance r1 is connected in series with a resistor of resistance R_{1}and a uniform metal wire of total resistance R_{2}.
A second cell of e.m.f. E

_{2}and internal resistance r_{2}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 I

_{1}, I_{2}and I_{3}, [1]
(ii) between E

_{1}, R_{1}, R_{2}, r_{1}, I_{1}and I_{2}in loop HBJFGH, [1]
(iii) between E

_{1}, E_{2}, r_{1}, r_{2}, R_{1}, R_{2}, I_{1}and I_{3}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) I

_{2}= I_{1}+ I_{3}(Kirchoff’s 1^{st}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 I

_{1}is coming from the +ve terminal of the cell, the same current should come back to (negative terminal of) the cell. So, current I_{1}flows through r_{1}and R_{1}which are in series.
At junction B, the current
I

_{1}changes into I_{2}and I_{3}(according to the above equation) but since we are considering loop HBJFGH, we can neglect I_{3}. From B to J, the current I_{2}flows. Since J is halfway on the wire, half of the total resistance of the wire (R_{2}) should be considered [since resistance is directly proportional to the length of wire –> R = ρL / A]. So, the p.d. across BJ = I_{2}(R_{2}/2).
Now, consider from the
junction J to part F, then G up to the negative terminal of cell E

_{1}. 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, I_{1}). And again, since J is halfway on the wire, part JF would correspond to a resistance of R_{2}/2. So, the p.d. across JF = I_{1}(R_{2}/2). Hence, the equation below is obtained.}
E

_{1}= I_{2}(R_{2}/2) + I_{1}(R_{2}/2) + I_{1}R_{1}+ I_{1}r_{1}
(iii)

{Tip: Put emf of cells (E

_{1}and E_{2}) on one sides of equation and potential differences in components on other side. (Kirchoff’s 2^{nd}law)
E

_{2}is in opposite direction of E_{1}(check direction of currents I_{3}and I_{1}).
I

_{2}not present since we are not considering BJ. Current coming from J towards F is I_{1}(corresponding to pd = I_{1}R_{2}/2). Current across r_{2}in JDCB is I_{3}(corresponding to pd = I_{3}r_{2}). In GH, pd = I_{1}R_{1}+ I_{1}r_{1}.}
E

_{1}– E_{2}= -I_{3}r_{2}+ I_{1}(R_{1}+ r_{1}+ R_{2}/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 E

_{2}.

{Since V = IR, current I =
V / R. As there is a difference in V, the current also changes}

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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).

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