The bridge: arms, diagonals, names
The Wheatstone bridge, NCERT tells us, is the circuit shown in Fig. 3.18 — an arrangement of four resistors that lets us compare resistances with great precision. Drawing it as a diamond clarifies its symmetry. Across one pair of diagonally opposite points (A and C) a source is connected; this diagonal (AC) is called the battery arm. Between the other two vertices, B and D, a galvanometer G — a device to detect currents — is connected. This line, BD, is called the galvanometer arm.
NCERT labels the four resistors $R_1, R_2, R_3, R_4$. In the equally common P–Q–R–S notation used by NIOS and many problems, the upper-left and upper-right arms are $P$ and $Q$, the lower-left and lower-right arms are $R$ and $S$, and the galvanometer sits between the two midpoints. The two notations describe the same circuit; we will keep both in view because NEET stems switch between them freely.
For simplicity NCERT assumes the cell has no internal resistance. In general there will be currents flowing across all the resistors as well as a current $I_g$ through G. Of special interest is the case of a balanced bridge, where the resistors are such that $I_g = 0$. Everything practical about the instrument flows from that one condition.
Deriving the balance condition
We can easily get the balance condition — the state in which there is no current through G — directly from Kirchhoff's rules. NCERT applies the junction rule first. With $I_g = 0$, Kirchhoff's junction rule applied to junctions D and B immediately gives the relations $I_1 = I_3$ and $I_2 = I_4$: whatever enters an arm passes straight through to the next, because none of it leaks across the galvanometer.
Next we apply Kirchhoff's loop rule to the two closed loops on either side of the galvanometer. The first loop, ADBA, gives (with $I_g = 0$):
$$-I_1 R_1 + 0 + I_2 R_2 = 0$$
and the second loop, CBDC, gives, upon using $I_3 = I_1$ and $I_4 = I_2$:
$$I_2 R_4 + 0 - I_1 R_3 = 0$$
From the first equation $\dfrac{I_2}{I_1} = \dfrac{R_1}{R_2}$, and from the second $\dfrac{I_2}{I_1} = \dfrac{R_3}{R_4}$. Equating the two ratios gives the result NCERT records as Eq. 3.64(a):
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$ In P–Q–R–S form this is the familiar $\dfrac{P}{Q} = \dfrac{R}{S}$. NCERT: "This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection."
Read physically, balance means points B and D are at the same potential, $V_B = V_D$. With no potential difference across the galvanometer arm, no current crosses it — the deflection is null. The condition can also be stated as "the products of opposite arms are equal," $P\,S = Q\,R$, which is the form most convenient when checking whether a given network is balanced.
| Quantity at balance | Value / relation | Reason |
|---|---|---|
| Galvanometer current $I_g$ | 0 (null) | Defining condition of balance |
| Potentials of B and D | $V_B = V_D$ | No drive across the galvanometer arm |
| Arm currents | $I_1 = I_3,\; I_2 = I_4$ | Junction rule with $I_g = 0$ |
| Balance condition | $\dfrac{P}{Q} = \dfrac{R}{S}$ i.e. $\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}$ | Loop rule on ADBA and CBDC |
| Equivalent form | $P\,S = Q\,R$ (opposite arms) | Cross-multiplying the balance ratio |
P/Q = R/S, not P/R = Q/S
The balance ratio pairs each arm with the one in the same branch on the same side of the galvanometer: $P/Q$ on top against $R/S$ on the bottom. Equivalently, opposite arms multiply equally, $PS = QR$. Students who write $P/R = Q/S$ have crossed the wrong way and will mis-place the unknown.
Safe check: cross-multiply to $PS = QR$. If swapping the galvanometer and battery diagonals leaves the same equation, you have it right — the bridge is symmetric in that exchange.
What balance is independent of
The power of the bridge is in what the balance condition does not contain. The relation $P/Q = R/S$ involves only the four resistances. NIOS states the two consequences directly. First, the balance condition at the null position is independent of the applied voltage — "even if you change the e.m.f of the cell, the balance condition will not change." Second, the measurement does not depend on the accuracy of calibration of the galvanometer, because the galvanometer is used only as a null indicator, a current detector, not as a measuring meter.
Both follow from $V_B = V_D$. Since there is no potential difference across the galvanometer arm at balance, the current through it is exactly zero whatever its resistance happens to be — a sensitive galvanometer and a sluggish one both read null at the same arm ratio. And since the EMF only scales the branch currents up or down without disturbing the ratio of potentials at B and D, the balance point sits at the same resistance ratio regardless of cell strength.
"Galvanometer reads zero" ≠ "galvanometer doesn't matter only at balance"
The independence from galvanometer resistance and from EMF holds at balance. Away from balance, both the galvanometer resistance and the EMF strongly affect how much current flows through G. A stem asking for the off-balance current (like NCERT's Example 3.7, where $I_g = 4.87$ mA through a 15 Ω galvanometer) needs the full Kirchhoff solution — the independence shortcut does not apply there.
If $I_g = 0$ is given or implied: use $P/Q = R/S$ and ignore G and EMF. If $I_g \neq 0$: set up loop equations for all branches.
Measuring an unknown resistance
The Wheatstone bridge and its balance condition provide a practical method for the determination of an unknown resistance. Following NCERT: suppose we have an unknown resistance, which we insert in the fourth arm, so $R_4$ is not known. Keeping known resistances $R_1$ and $R_2$ in the first and second arms, we go on varying $R_3$ until the galvanometer shows a null deflection. The bridge is then balanced, and from the balance condition the value of the unknown resistance is given by Eq. 3.64(b):
$$R_4 = R_3\,\frac{R_2}{R_1}$$
In P–Q–R–S notation, with the unknown in arm $S$, NIOS writes the same result as $S = \dfrac{Q R}{P}$. The known ratio arms $P$ and $Q$ set the scale, the variable arm $R$ is tuned to balance, and the unknown follows by simple proportion. Because the answer depends only on a ratio of known resistances and a null reading, it can be made very accurate.
Every step of the balance derivation rests on the junction and loop rules. Revisit Kirchhoff's rules if the loop signs feel shaky.
The metre bridge (slide-wire)
A practical device using this principle is called the metre bridge (also written meter bridge). Here two of the four arms are replaced by a single uniform resistance wire exactly one metre long, stretched along a metre scale. The unknown resistance $R$ goes in one gap and a known standard $S$ in the other gap; the cell drives current through the wire, and a sliding contact called a jockey taps along the wire feeding the galvanometer.
Because the wire is uniform, the resistance of each segment is proportional to its length. If the null point is found at a distance $l$ (in cm) from end A, the left segment has resistance proportional to $l$ and the right segment proportional to $(100 - l)$. These two segments play the role of the ratio arms, so the balance condition $\dfrac{R}{S} = \dfrac{l}{100 - l}$ rearranges to give the unknown directly:
$$R = S\;\frac{l}{100 - l}$$
The standard $S$ is chosen so that the balance point falls near the middle of the wire, where, as we will see, the bridge is most sensitive and the reading is most reliable. Only a ratio of lengths and one known resistance enter the result — exactly the kind of robust, calibration-free measurement that the balance condition guarantees.
In a metre bridge, an unknown resistance balances a standard $S = 6\ \Omega$ at a null point $40$ cm from the left end. Find the unknown resistance $R$.
With $l = 40$ cm, $100 - l = 60$ cm and the unknown in the left gap:
$$R = S\,\frac{l}{100 - l} = 6 \times \frac{40}{60} = 4\ \Omega.$$
The balance point lies left of centre because the unknown (4 Ω) is smaller than the standard (6 Ω). Had it landed near 50 cm, the two resistances would have been nearly equal — the most sensitive setting.
Bridge sensitivity
Accuracy in a Wheatstone bridge depends on its sensitivity — how sharply the galvanometer responds to a small departure from the exact balance point. The more strongly G deflects for a tiny mismatch in arm ratio, the more precisely the balance can be located. NIOS records the practical rule: the bridge has the greatest sensitivity when the resistances in all the arms are nearly equal.
This is why, for the most precise measurement, the ratio arms $P$ and $Q$ are kept approximately equal (and not very large), and in a metre bridge the standard is chosen to bring the null point near the middle of the wire. Equal arms also keep the result clear of the systematic errors that creep in when one arm dominates. The single design lesson is short: balance is set by the ratio; precision is set by keeping the arms comparable.
Wheatstone bridge in one screen
- Four resistors in a diamond; battery across AC, galvanometer across BD.
- Balance: $I_g = 0$ when $\dfrac{P}{Q} = \dfrac{R}{S}$, i.e. $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$, equivalently $PS = QR$.
- At balance $V_B = V_D$; the null reading is independent of the galvanometer resistance and of the cell EMF.
- Unknown by bridge: $R_4 = R_3\,R_2/R_1$, i.e. $S = QR/P$.
- Metre bridge: $R = S\,\dfrac{l}{100 - l}$ with $l$ in cm.
- Greatest sensitivity (most precise) when all four arms are nearly equal.
- A balanced bridge lets you drop the galvanometer arm when finding $R_{eq}$.