A potentiometer circuit that is used as a means of comparing potential differences

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 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 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 2^nd 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}