The post office box is beyond the Class 12 syllabus. It appears in JEE problems and in older practical manuals, and it rests entirely on the balance of the Wheatstone bridge from the last lesson.
A bridge in a box
A post office box is a Wheatstone bridge packed into one wooden box. Its name comes from the telegraph engineers of the British Post Office, who used such boxes to test lines. Brass blocks on its lid are joined by resistance coils. Pulling a brass plug out of the gap between two blocks puts that coil in circuit.
Two arms are the ratio arms, \(P\) and \(Q\). Each can be set to \(10\), \(100\) or \(1000\,\Omega\). The third arm \(R\) is a resistance box that can be set in whole ohms up to several thousand ohms. The unknown \(X\) is joined as the fourth arm. Two tap keys complete it: the battery key \(K_1\) and the galvanometer key \(K_2\).
With \(P\) and \(Q\) in one branch and \(R\) and \(X\) in the other, balance needs \(P/Q = R/X\). So the unknown is \(X = RQ/P\). The ratio \(Q/P\) is a multiplier chosen by the student; \(R\) is what the student adjusts.
The procedure, stage by stage
First set \(P = Q = 10\,\Omega\), so \(X = R\). Test the connections with two extreme settings. With \(R = 0\) the galvanometer should kick one way. With the infinity plug out, so that \(R\) is infinite, it should kick the other way. Opposite kicks show that a balance lies somewhere between.
Now change \(R\) in steps of \(1\,\Omega\). Suppose the kick is to the left at \(27\,\Omega\) and to the right at \(28\,\Omega\). The null lies between, so \(27 < X < 28\,\Omega\). The box moves only in whole ohms, so this is all the first stage can say.
Then set \(P = 100\,\Omega\) and keep \(Q = 10\,\Omega\), so \(X = R/10\). Now \(R\) must be near \(270\) to \(280\,\Omega\). Suppose the kick reverses between \(273\) and \(274\,\Omega\). Then \(27.3 < X < 27.4\,\Omega\). Finally set \(P = 1000\,\Omega\), so \(X = R/100\). If the needle rests at \(R = 2734\,\Omega\), then \(X = 27.34\,\Omega\).
In the figure above the needle leans right when \(RQ/P\) is larger than \(X\) and left when it is smaller. Try each chip. With \(P = 10\,\Omega\) the dot on the number line jumps a whole ohm at a time. With \(P = 1000\,\Omega\) it moves a hundredth of an ohm at a time, and it can land on \(27.34\,\Omega\) itself.
A post office box has \(Q = 10\,\Omega\). With \(P = 10\,\Omega\) the kick reverses between \(R = 56\) and \(57\,\Omega\). With \(P = 100\,\Omega\) it reverses between \(562\) and \(563\,\Omega\). With \(P = 1000\,\Omega\) the needle rests at \(5627\,\Omega\). What does each stage say about \(X\)?
At each stage \(X = RQ/P\), with \(Q = 10\,\Omega\).
Stage 1: \(X = R\). Stage 2: \(X = R/10\). Stage 3: \(X = R/100\).
Stage 1: \(56 < X < 57\,\Omega\). Stage 2: \(56.2 < X < 56.3\,\Omega\). Stage 3: \(X = 5627/100 = 56.27\,\Omega\).
The unknown is \(56.27\,\Omega\), found to a hundredth of an ohm with a box that moves only in whole ohms.
A box that moves only in whole ohms has found a resistance to a hundredth of an ohm. Values used in this lesson: the ratio arms of \(10\), \(100\) and \(1000\,\Omega\), and the settings of the workbook's second sheet.
In a post office box \(P = 1000\,\Omega\) and \(Q = 10\,\Omega\). The galvanometer shows no deflection when the resistance box reads \(4719\,\Omega\). Find \(X\), in ohms.
Put the steps of measuring a resistance with a post office box in order.
One decimal place per stage
Why these three values? The resistance box moves only in steps of \(1\,\Omega\). With \(Q/P = 1\) a step in \(R\) is a step of \(1\,\Omega\) in \(X\). With \(Q/P = 1/10\) the same step is \(0.1\,\Omega\) in \(X\), and with \(Q/P = 1/100\) it is \(0.01\,\Omega\). Each stage adds one decimal place with the same box.
The ratio also reaches unknowns that the box alone could not. A very large \(X\), beyond the box's range, is measured with the ratio turned round: \(P = 10\,\Omega\) and \(Q = 1000\,\Omega\) give \(X = 100R\). So a box setting of \(50\,\Omega\) then balances \(5000\,\Omega\).
Why are the ratio arms of a post office box made \(10\), \(100\) and \(1000\,\Omega\)?
Battery key first
The coils of the box are wires wound on bobbins. A coil opposes a sudden change of the current through it, a property called self-inductance. It belongs to the chapter Eddy Currents, Generators and Inductance. For now it is enough to know its effect: when the battery key closes, the currents in the four arms take a moment to settle.
During that moment the arms do not share the voltage in their final ratio. B and D can differ in potential even for a bridge that will be balanced. If the galvanometer is already joined, it receives a jolt that has nothing to do with balance. So the battery key \(K_1\) is pressed first and the galvanometer key \(K_2\) after it. When the reading is done, \(K_2\) is released first.
In Fig 4.1 the steady deflection is the same in both chips. Only the throw differs. A student who presses \(K_2\) first may read the throw as a large imbalance and chase a wrong setting. The coils of good resistance boxes are wound doubled back on themselves to keep their inductance small, but not zero.
While using a post office box, why is the battery key pressed before the galvanometer key?
- Writing \(X = RP/Q\).Balance is \(P/Q = R/X\), so \(X = RQ/P\). With \(P = 1000\,\Omega\) and \(Q = 10\,\Omega\), the box reading is divided by \(100\).
- Jumping to \(P = 1000\,\Omega\) at once.The equal-arms stage must first bracket \(X\), or the search at the larger ratio has no starting range.
- Reading the first throw of the needle as the deflection.Press the battery key first. The steady deflection, not the throw, tells which way to change \(R\).