Measuring Instruments, Bridges and the Potentiometer – Direct Current Circuits (Senior 4 Physics) | Ganzaa

Direct Current Circuits

Lessons

Senior 4 · Physics

Direct Current Circuits

5 lessons

Lesson 5 of 5

Measuring Instruments, Bridges and the Potentiometer

25 min

Measuring without disturbing

Every meter you connect into a circuit changes it a little: an ammeter adds resistance to the branch it measures, and a voltmeter draws some current from the points it measures across. In lesson 3 you saw a poor voltmeter read 0.83 V where 5.0 V should have been. Physicists solved this problem in two ways: by designing meters that disturb the circuit as little as possible, and by inventing null methods, in which you adjust the circuit until a sensitive meter reads zero, so that at the moment of measurement no current flows through it at all.

The galvanometer, the ammeter and the voltmeter

A moving-coil galvanometer is a very sensitive current detector: a coil on a pivot turns in a magnetic field when a tiny current flows, moving a pointer. It is described by its full-scale deflection current $I_{g}$ (often a few milliamperes or less) and its coil resistance $R_{g}$ . The same galvanometer becomes an ammeter or a voltmeter of any range by adding a resistor (Figure 5.1).

Figure 5.1: Converting a galvanometer (full-scale current $I_{g}$ , resistance $R_{g}$ ).

Ammeter. A small shunt resistance $S$ in parallel carries most of the current. At full scale, the galvanometer carries $I_{g}$ and the shunt $I - I_{g}$ , with the same p.d. across both: $I_{g} R_{g} = (I - I_{g}) S$ , so

S = \frac{I_g R_g}{I - I_g}

An ammeter is connected in series and should have a very low resistance, so that it hardly changes the current it measures; an ideal ammeter has zero resistance.

Voltmeter. A large multiplier resistance $R_{m}$ in series limits the current to $I_{g}$ at the full-scale p.d. $V$ : $V = I_{g} (R_{g} + R_{m})$ , so

R_m = \frac{V}{I_g} - R_g

A voltmeter is connected in parallel and should have a very high resistance, much higher than the resistance it is connected across; an ideal voltmeter draws no current.

Worked example 1: one galvanometer, two meters

A galvanometer has a full-scale deflection of 1.0 mA and a resistance of $50\,\Omega$ . How can it be converted into (a) an ammeter reading up to 1.0 A, (b) a voltmeter reading up to 10 V?

\text{(a)}\;\; S = \frac{1.0 \times 10^{-3} \times 50}{1.0 - 0.0010} = 0.050\,\Omega \ \text{in parallel} \qquad \text{(b)}\;\; R_m = \frac{10}{1.0 \times 10^{-3}} - 50 = 9950\,\Omega \ \text{in series}

The ammeter's total resistance is about $0.050\,\Omega$ ; the voltmeter's is $10\,000\,\Omega$ .

The Wheatstone bridge

Figure 5.2: The Wheatstone bridge: four resistors and a galvanometer.

Four resistors P, Q, R and S are connected in a diamond, with a cell across A and C and a galvanometer across B and D (Figure 5.2). One of the resistors is adjusted until the galvanometer reads zero: the bridge is then balanced. With no current through G, B and D are at the same potential, so the current $I_{1}$ is the same in P and Q, and $I_{2}$ is the same in R and S. Equal p.d.s from A to B and from A to D, and from B to C and from D to C, give

I_1P = I_2R \quad\text{and}\quad I_1Q = I_2S \quad\Rightarrow\quad \frac{P}{Q} = \frac{R}{S}

Three known resistances give the fourth. Because nothing flows through the galvanometer at balance, its resistance and calibration do not matter, and neither do small changes in the cell's e.m.f.: this is a null method.

The metre bridge

In a metre bridge (slide-wire bridge), the lower two arms are the two parts of a uniform wire 1.00 m long, split by a sliding contact called a jockey (Figure 5.3). A uniform wire has resistance proportional to length, so at balance

\frac{X}{R} = \frac{l}{100 - l} \qquad (l \text{ in cm})

Figure 5.3: A metre bridge. The jockey is tapped along the wire until the galvanometer shows no deflection.

The result is most precise when the balance point is near the middle of the wire, so choose $R$ close to the expected $X$ . Repeating with X and R interchanged and averaging cancels small end errors.

Definition

A null method is a measurement in which the circuit is adjusted until a detector reads zero; the result then depends only on known quantities, not on the detector's accuracy. A Wheatstone bridge is balanced when $\frac{P}{Q} = \frac{R}{S}$ . A potentiometer is a uniform resistance wire carrying a steady current, used to measure or compare p.d.s and e.m.f.s by balancing them against part of the wire. Example: a metre bridge with $R = 10\,\Omega$ balanced at $l = 40.0\,\mathrm{cm}$ gives $X = 6.67\,\Omega$ .

Worked example 2: a metre bridge

An unknown resistor X is in the left gap and a $10.0\,\Omega$ resistor in the right gap. Balance is found 40.0 cm from the left end. Find X.

X = R\,\frac{l}{100 - l} = 10.0 \times \frac{40.0}{60.0} = 6.67\,\Omega

With the resistors swapped, balance should occur at 60.0 cm; any difference reveals an end error to be averaged out.

The potentiometer

Figure 5.4: A potentiometer comparing e.m.f.s.

A driver cell sends a steady current along a uniform wire AB, so the potential falls uniformly along it: the p.d. between A and a point at distance $l$ is $k l$ , where $k$ is the potential gradient (volts per metre). A cell under test is connected, positive terminal towards A, through a galvanometer to a jockey (Figure 5.4). At the balance length $l_{1}$ , no current flows through the test cell, so the p.d. across $l_{1}$ of wire equals its full e.m.f.:

E_{1} = k l_{1}

Comparing e.m.f.s. Replace the cell by a second cell $E_{2}$ (for example a standard cell of known e.m.f.) and find its balance length $l_{2}$ :

\frac{E_1}{E_2} = \frac{l_1}{l_2}

Internal resistance. Find the balance length $l_{0}$ for the cell on open circuit ( $E = k l_{0}$ ); then connect a known resistor $R$ across the cell and find the new balance length $l$ (now measuring the terminal p.d. $V = k l$ ). Since $\frac{E}{V} = \frac{R + r}{R}$ :

r = R\,\frac{l_0 - l}{l}

A voltmeter connected across a cell draws current and so reads less than the e.m.f.; the potentiometer draws none at balance and measures the e.m.f. itself.

Worked example 3: a potentiometer

(a) A standard cell of e.m.f. 1.018 V balances at 50.9 cm, and a dry cell at 75.0 cm on the same wire. Find the e.m.f. of the dry cell. (b) With a $10.0\,\Omega$ resistor across the dry cell, balance moves to 68.2 cm. Find its internal resistance.

\text{(a)}\;\; E_1 = 1.018 \times \frac{75.0}{50.9} = 1.50\,\mathrm{V} \qquad \text{(b)}\;\; r = 10.0 \times \frac{75.0 - 68.2}{68.2} = 1.0\,\Omega

No balance point can be found if the driver cell's e.m.f. is smaller than the e.m.f. being measured, or if the test cell is connected the wrong way round.

Experiment 5.1: measuring a resistance with a metre bridge

Apparatus: a metre bridge, a resistance box (or standard resistors), the unknown resistor (for example a coil of wire), a centre-zero galvanometer, a cell, a jockey, leads.

Connect the circuit of Figure 5.3. Choose $R$ close to the expected value of $X$ .
Touch the jockey briefly on the wire near each end: the galvanometer should deflect in opposite directions. Then find the point of zero deflection by tapping, not sliding (sliding scrapes the wire and changes its uniformity).
Record $l$ , calculate X, then interchange X and R and repeat. Repeat for two other values of R and average.
If X is a length of wire, measure its length and diameter and find the resistivity $\rho = \frac{XA}{L}$ .

Warning

Safety and care. Use a low-voltage cell, and close the circuit only while taking a reading so the wire does not heat (heating changes its resistance). Protect the galvanometer: start with a high resistance or a shunt in series with it, and remove it only near balance.

Warning

Common mistakes. (1) Putting a shunt in series or a multiplier in parallel. (2) Connecting an ammeter in parallel across a component: its low resistance acts almost as a short circuit and can damage it. (3) Forgetting that a metre-bridge length must be measured from the end next to the resistor it belongs to. (4) Connecting the test cell in a potentiometer with its positive terminal towards the wrong end, so that no balance point exists.

Tip

"Ammeters: all the current passes through, so low resistance, in series. Voltmeters: very high resistance, across the component." For bridges and potentiometers, think "balance means no current through the detector".

The metre bridge: principle and working (Khan Academy India). Watch how the balanced Wheatstone-bridge condition becomes a ratio of two lengths of uniform wire.

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Practice

A galvanometer has a full-scale current of 2.0 mA and a resistance of $25\,\Omega$ . Find the shunt needed to make a 0–5.0 A ammeter and the multiplier needed for a 0–20 V voltmeter.
Explain why an ammeter should have a low resistance and a voltmeter a high resistance.
In a Wheatstone bridge, $P = 100\,\Omega$ , $Q = 10\,\Omega$ and balance occurs when $R = 470\,\Omega$ . Find S.
A metre bridge balances at 62.5 cm from the end next to an unknown resistor X when a $5.0\,\Omega$ resistor is in the other gap. Find X.
A potentiometer wire 1.00 m long carries a steady current that gives a p.d. of 2.00 V across it. Find the potential gradient and the balance length for a 1.50 V cell.
A cell balances at 72.0 cm on open circuit and at 60.0 cm when a $5.0\,\Omega$ resistor is connected across it. Find its internal resistance.
A student measures the e.m.f. of a cell with a voltmeter of resistance $200\,\Omega$ and gets 1.40 V; a potentiometer gives 1.50 V. Evaluate the two results and use them to estimate the cell's internal resistance.

Show answersHide answers

$S = \frac{0.0020 \times 25}{5.0 - 0.0020} = 0.010\,\Omega$ ; $R_m = \frac{20}{0.0020} - 25 = 9975\,\Omega \approx 10\,\mathrm{k\Omega}$ .
An ammeter is in series, so its resistance adds to the circuit; a low value hardly changes the current. A voltmeter is in parallel, so it draws current; a high value draws very little and hardly changes the p.d. it measures.
$\frac{P}{Q} = \frac{R}{S} \Rightarrow S = R\,\frac{Q}{P} = 470 \times \frac{10}{100} = 47\,\Omega$ .
$X = 5.0 \times \frac{62.5}{37.5} = 8.3\,\Omega$ .
$k = 2.00\,\mathrm{V\,m^{-1}}$ ; $l = \frac{1.50}{2.00} = 0.750\,\mathrm{m}$ .
$r = 5.0 \times \frac{72.0 - 60.0}{60.0} = 1.0\,\Omega$ .
The potentiometer value, 1.50 V, is the true e.m.f., because no current flows at balance. The voltmeter draws current, so it reads the terminal p.d.: $1.40 = 1.50 - I r$ with $I = \frac{1.40}{200} = 7.0\,\mathrm{mA}$ , giving $r = \frac{0.10}{0.0070} = 14\,\Omega$ . (Such a large $r$ suggests an old cell.) The voltmeter's result is not wrong as a terminal p.d., but it is not the e.m.f.

Quick check

How is a galvanometer converted into an ammeter?

Key points

Ammeter = galvanometer + small shunt in parallel, $S = \frac{I_gR_g}{I - I_g}$ ; voltmeter = galvanometer + large multiplier in series, $R_m = \frac{V}{I_g} - R_g$ .
Ideal ammeters have zero resistance; ideal voltmeters infinite resistance.
Wheatstone bridge at balance: $\frac{P}{Q} = \frac{R}{S}$ ; metre bridge: $\frac{X}{R} = \frac{l}{100 - l}$ .
Potentiometer: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$ , $r = R\frac{l_0 - l}{l}$ ; null methods draw no current at balance.

1Measuring without disturbing

2The galvanometer, the ammeter and the voltmeter

3The Wheatstone bridge