13.8.1 Basic Circuit

Null Deflection

An archaic application of the potential divider is to use a slide wire potentiometer to measure voltages. It is a lot like how we would use a lever balance to measure the weight (alright mass, if you’re particular) of a teddy bear. We put the teddy bear on the left. We adjust the number of standard weights on the right until the lever is balanced… you know how it works.

Time to get back to our slide wire potentiometer (the “voltage” balance). Let’s start with the most basic setup shown below.

First, notice the long thin uniform wire (aka the slide wire) and the cell of known emf E₁ (aka the driver cell) connected across it. This section of the circuit is called the driver circuit. Next, notice that there is another cell E₂. This is the cell with the unknown emf we are trying to measure.

Now notice that both terminals of E₂ are connected to two points on the slide wire: C is connected to A by a direct wire; D is connected to X via a galvanometer and jockey. This make V_CD and V_AX the two voltages we are seeking to balance.

In other words, V_CD is the “teddy bear”, and V_AX is the “standard weights”. Sliding the jockey along the slide wire is analogous to “adjusting the standard weights on the lever balance”. So we slide the jockey along the slide wire (i.e. vary the position of X and thus V_AX) until we locate the point X where the galvanometer shows null (i.e. zero) deflection. This magic point X is called the null point. The length AX is called the balance length.

Now what is V_AX? Analysing the driver circuit using the potential divider principle, we can compute that $\displaystyle {{V}_{{AX}}}=\frac{{AX}}{{AB}}\times {{E}_{1}}$ . What is V_CD? For this simple example, V_CD is simply E₂. So

$\displaystyle \begin{aligned}{{V}_{{CD}}}&={{V}_{{AX}}}\\\text{ }{{E}_{2}}&=\frac{{AX}}{{AB}}\times {{E}_{1}}\end{aligned}$

Example

The potentiometer below uses a driver cell of 6.0 V and a uniform slide wire of 80 cm. Calculate E₂, if the balance length AX is found to be 24 cm.

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Solution

Since

$\displaystyle \begin{aligned}{{V}_{{CD}}}&={{V}_{{AX}}}\\{{E}_{2}}&=\frac{{24}}{{80}}(6.0)&=1.8\text{ V}\end{aligned}$

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Got Deflection

Consider the potentiometer below, where a 6.0 V driver cell and an 80 cm uniform slide wire is used to balance the voltage of a 3.0 V cell.

First, let me highlight this fact: any current i running from C to A must return as i from X to D. So if the galvanometer indicates that there is no current in XD, it also means that there is no current in CA. That’s why we only need one galvanometer to tell us whether there is any current running between E₂ and the driver circuit.

Secondly, once C and D are connected to A and X respectively, V_AX and V_CD will become equal (assuming the connecting wires and galvanometer have zero resistance). What’s crucial is whether a current between E₂ and the primary circuit is required to achieve this equality.

If the jockey connects to point X’ such that that $\displaystyle AX'=40\text{ cm}$ then obviously $\displaystyle {{V}_{{AX'}}}={{V}_{{CD}}}$ even before the connection is made. So completing the connection does not affect anything in the driver circuit. E₂ will neither add any current to, nor divert any current from, the slide wire (because this will break $\displaystyle {{V}_{{AX'}}}={{V}_{{CD}}}$ ). This is why the galvanometer shows zero deflection. Not having any current between E₂ and the primary circuit also allows us to continue to calculate the currents and voltages in the driver circuit as if it is detached from E₂.

AX is too short

Now what if the jockey connects to a point X left of X’ so that $\displaystyle AX<40\text{ cm}$ ? Notice that V_AX is less than V_CD before the connection is made. So making the connection must affect something in the driver circuit. Because if nothing changes, V_AX will remain smaller than V_CD! So how does V_AX become equal to V_CD? E₂ must drive a current i from C to A into the slide wire, so as to increase the current in AX in order for V_AX to increase towards $\displaystyle {{V}_{{CD}}}=3.0\text{ V}$ . As i returns to E₂ from X to D, it is detected by the galvanometer. This is why the galvanometer shows a deflection.

AX is too long

Now what if the jockey connects to a point X right of X’ so that $\displaystyle AX>40\text{ cm}$ ? Notice that V_AX is more than V_CD before the connection is made. So making the connection must affect something in the driver circuit. Because if nothing changes, V_AX will remain larger than V_CD! So what happen this time? E₂ must accept a current i from A to C, so as to reduce the current in AX in order for V_AX to decrease towards $\displaystyle {{V}_{{CD}}}=3.0\text{ V}$ . As i returns to the slide wire from D to X, it is detected by the galvanometer. This time, the galvanometer will show a deflection in the other direction.