In the physics laboratory of a Class 12 practical, a student slides a small brass jockey along a bare wire stretched over a metre scale. At each touch the needle of a galvanometer kicks one way or the other. Near one point the kick shrinks to nothing, and the needle rests. The student notes the reading on the scale. From that one length, and a known resistance, comes the value of an unknown resistance.
Four resistors and a galvanometer
A Wheatstone bridge has four resistors in a diamond. \(P\) runs from A to B and \(Q\) from B to C; \(R\) runs from A to D and \(S\) from D to C. A cell joins A and C. A galvanometer, a sensitive current detector, joins the middle points B and D. The galvanometer asks one question: are B and D at the same potential?
Take C as zero and remove the galvanometer for a moment. The branch ABC has \(5\,\Omega\) on \(10\,\text{V}\), so it carries \(2\,\text{A}\), and B sits \(2 \times 3 = 6\,\text{V}\) above C. With \(S = 8\,\Omega\) the branch ADC has \(12\,\Omega\) and carries \(10/12\,\text{A}\), so D sits at \(8 \times 10/12 = 6.67\,\text{V}\). D is higher than B. Put the galvanometer back, and current flows through it from D to B.
Take the chip \(S = 6\,\Omega\). Now D sits at \(6 \times 10/10 = 6\,\text{V}\), the same as B. No potential difference drives the galvanometer, and it carries no current. The bridge is balanced. In each branch the upper arm takes the same fraction of the cell's voltage: \(P/(P + Q) = R/(R + S)\), which is \(P/Q = R/S\).
\[ I_G = 0 \Rightarrow V_B = V_D \]
Why. at balance the galvanometer carries no current, so no potential difference lies across it.
\[ I_1 P = I_2 R \]
Why. \(I_1\) flows through \(P\) and \(I_2\) through \(R\); both start at A and fall by the same amount to B and to D.
\[ I_1 Q = I_2 S \]
Why. with no current into the galvanometer, the same \(I_1\) continues through \(Q\) and \(I_2\) through \(S\), and both fall from B and D to C.
\[ \dfrac{P}{Q} = \dfrac{R}{S} \]
Read it. divide step 2 by step 3; the currents cancel.
A Wheatstone bridge with arms \(P\) and \(Q\) in one branch and \(R\) and \(S\) in the other is balanced when \(P/Q = R/S\). The two middle points are then at the same potential, and the galvanometer between them carries no current.
Balance has three consequences. The galvanometer carried nothing, so removing it changes no current in any arm. Its own resistance does not matter either, so a resistance in series with it does not upset the balance. And the condition can be written \(P/R = Q/S\), the balance of the same bridge with the cell and the galvanometer swapped. So the swapped bridge is balanced too.
A bridge that is nearly balanced is set right by one arm. With \(P = Q\), balance needs \(R = S\). If \(R = 5\,\Omega\) and \(S = 5.2\,\Omega\), either raise \(R\) by \(0.2\,\Omega\) or lower \(S\) by \(0.2\,\Omega\). In practice the adjustable arm is changed.
In a Wheatstone bridge \(P = 10\,\Omega\), \(Q = 20\,\Omega\) and \(R = 15\,\Omega\). What value of \(S\) balances it?
Balance needs \(P/Q = R/S\), so \(S = RQ/P = 15 \times 20/10 = 30\,\Omega\).
Check the ratios: \(10/20 = 15/30 = 1/2\).
Values used in this lesson: the bridges of the workbook's second sheet and a meter bridge wire \(100\,\text{cm}\) long.
A Wheatstone bridge is balanced. The galvanometer is taken out of the circuit. What happens to the currents in the four arms?
One wire for two arms
A meter bridge is a Wheatstone bridge built for measurement. A uniform wire \(100\,\text{cm}\) long, usually of manganin or constantan, is stretched along a metre scale. Thick copper strips at its ends make two gaps. The unknown resistance \(X\) goes in the left gap and a known resistance \(R\) from a resistance box in the right one. A cell drives current along the wire, and a galvanometer joins the strip between the gaps to a sliding contact, the jockey.
The jockey divides the wire into two lengths, \(l\) and \(100 - l\). The wire is uniform, so the resistance of each part is proportional to its length: \(\rho l/A\) and \(\rho(100 - l)/A\). These are the two lower arms. At the null point the galvanometer rests, and the balance condition gives \(X/R = l/(100 - l)\).
In the figure above, \(X = 4\,\Omega\) balances at \(40\,\text{cm}\): \(4/6 = 40/60\). Swap the two, and the null moves to \(60\,\text{cm}\), the mirror image about the middle. Short of the null the needle leans left; beyond it the needle leans right. The driver's positive terminal is at the left end. Short of the null, current flows up from the wire to D; beyond it, down from D to the wire.
A meter bridge with \(R = 5\,\Omega\) balances at \(50\,\text{cm}\). Find \(X\). A \(5\,\Omega\) resistor is then joined in parallel with \(X\). Find the new null point.
The null at the middle means the two gaps hold equal resistances.
\(X = R\,l/(100 - l)\). With the shunt, the left gap holds \(X \parallel 5\,\Omega\).
\(X = 5 \times 50/50 = 5\,\Omega\). The left gap now holds \(2.5\,\Omega\): \(2.5/5 = l/(100 - l)\), so \(100 - l = 2l\) and \(l = 33.3\,\text{cm}\).
\(X = 5\,\Omega\), and with the shunt the null moves to \(33.3\,\text{cm}\).
A meter bridge has a known \(12\,\Omega\) in the right gap and an unknown \(X\) in the left gap. The null point is \(40\,\text{cm}\) from the left end. Find \(X\), in ohms.
Two resistances \(R_1\) and \(R_2\), with \(R_2 > R_1\), in the two gaps of a meter bridge balance at \(1/3\,\text{m}\) from the end next to \(R_1\). A \(6\,\Omega\) resistor is added in series with \(R_1\), and the null moves to \(2/3\,\text{m}\) from the same end. Find \(R_1\), in ohms.
The bridge was first described in 1833 by Samuel Hunter Christie, a mathematician in England. Few noticed it. Ten years later Charles Wheatstone, a pioneer of the electric telegraph, described it to the Royal Society. He gave Christie the credit. The name stuck to Wheatstone anyway. Telegraph engineers soon used such bridges to test their long lines.
Now the practical of the opening. Say the student takes \(R = 3\,\Omega\) from the box and finds the null at \(40.0\,\text{cm}\). Then \(X = 3 \times 40/60 = 2.00\,\Omega\). The needle rests at one point because only there are the two middle points at one potential. A millimetre of error in the null changes \(X\) by only about \(0.008\,\Omega\), as the next topic shows.
End corrections
The formula assumes that the arms start exactly at the zero and the \(100\) of the scale. In a real bridge the wire's ends lie a little beyond the marks, under the clamps. The joints and strips add a little resistance too. These add small lengths \(\alpha\) at the left end and \(\beta\) at the right, the end corrections: \(P/Q = (l + \alpha)/(100 - l + \beta)\).
Two balances find them. With \(2\,\Omega\) in the left gap and \(3\,\Omega\) in the right, the null is at \(40.1\,\text{cm}\). Swapped, it is at \(60.4\,\text{cm}\). The first gives \(2(59.9 + \beta) = 3(40.1 + \alpha)\). The second gives \(3(39.6 + \beta) = 2(60.4 + \alpha)\). Together they give \(\alpha = 0.5\,\text{cm}\) and \(\beta = 1\,\text{cm}\).
Why the null belongs near the middle
The null point can be read only to about a millimetre. Write \(X = R\,l/(100 - l)\) and take logarithms before differentiating. A small error \(\delta l\) gives a fractional error \(\delta X/X = \delta l\,(1/l + 1/(100 - l))\). The bracket is least when \(l = 50\,\text{cm}\).
In the figure above a millimetre costs \(0.4\,\%\) at \(50\,\text{cm}\), but \(1.1\,\%\) at \(10\,\text{cm}\) and \(2.1\,\%\) at \(5\,\text{cm}\). So the student chooses \(R\) close to \(X\), which puts the null near the middle. The bridge is also most sensitive when all four arms are of similar size.
What moves the null, and what does not
A high resistance in series with the galvanometer does not move the null, because at the null no current flows through it. It keeps the galvanometer safe from large kicks while the student hunts. Near the null it is shorted out, so that the galvanometer is fully sensitive. The thick copper strips have almost no resistance, so the arms are only \(X\), \(R\) and the two parts of the wire.
Heat matters. If the wire warms evenly during a long experiment, every centimetre's resistance grows by the same factor. The ratio \(l/(100 - l)\) at balance is unchanged, so the null does not move. But if a copper coil \(X\) in the left gap warms, its resistance rises. Then \(l/(100 - l)\) must rise too, and the null moves to the right.
In a meter bridge the unknown \(X\), a copper coil in the left gap, warms up during the experiment. The known \(R\) and the wire stay at room temperature. What happens to the null point?
Why is the known resistance of a meter bridge chosen so that the null point falls near the middle of the wire?
The null point can be read only to about a millimetre. The fractional error in \(X\) is \(\delta l\) times \((1/l + 1/(100 - l))\), and this is smallest when \(l = 50\,\text{cm}\). Near either end one of the two lengths is small, so the same millimetre makes a much larger fractional error. Choosing \(R\) close to \(X\) puts the null near the middle and keeps the error least.
- Putting the length on the wrong side of the ratio.The length measured from the end next to \(X\) goes with \(X\): \(X/R = l/(100 - l)\).
- Thinking that the galvanometer's resistance, or a protective resistance with it, changes the balance.At balance no current flows in that branch, so its resistance does not enter the condition.
- Expecting a uniformly warming bridge wire to shift the null.Only the ratio of the two lengths' resistances matters, and even warming leaves it unchanged.
- Taking readings with the null near one end.Choose \(R\) near \(X\), so that the null falls near \(50\,\text{cm}\), where the fractional error is least.