Wheatstone Bridge — All Variants

Wheatstone Bridge — All Variants You Need

The Wheatstone bridge looks intimidating with its diamond shape and four resistors, but the core idea is one line: when the bridge is balanced, no current flows through the galvanometer. Learn that condition cold and 90% of bridge problems become arithmetic.

The bridge appears in JEE Main, JEE Advanced, NEET, and CBSE class 12 boards — usually as a circuit-analysis question, sometimes as a combination-of-resistances simplifier. We will cover the basic balance condition, the unbalanced case, the meter bridge variant, and the trickier networks where Wheatstone hides inside a larger circuit.

Key Terms & Definitions

Galvanometer. A sensitive ammeter used to detect very small currents. In the Wheatstone bridge, it sits across the central diagonal.

Balance condition. The state where the galvanometer reads zero — no current flows through it. This happens when the four resistor ratios are matched.

Bridge arms. The four resistors $P, Q, R, S$ arranged in a diamond. $P$ and $Q$ are typically on one side, $R$ and $S$ on the other.

Meter bridge. A practical version using a uniform resistance wire of length 1 m, with two known resistors and a sliding contact (jockey).

The Balance Condition

When the galvanometer reads zero current,

$\frac{P}{Q} = \frac{R}{S}$

Equivalently: $PS = QR$ .

This comes from applying Kirchhoff’s laws and demanding that the potential at the two galvanometer terminals is equal. When that holds, no current flows through the galvanometer, and the bridge “balances.”

A useful re-arrangement: if you know three resistances and the bridge balances, you can find the fourth: $S = QR/P$ .

How the Bridge Works — Derivation

When the galvanometer reads zero, no current flows through it. So the same current $I_1$ passes through $P$ and $Q$ , and the same current $I_2$ passes through $R$ and $S$ .

Let the bridge ends be $A$ (battery +) and $C$ (battery −). Galvanometer connects nodes $B$ (between $P, Q$ ) and $D$ (between $R, S$ ). Balance means $V_B = V_D$ .

$V_A - V_B = I_1 P$ and $V_A - V_D = I_2 R$ . Setting $V_B = V_D$ : $I_1 P = I_2 R$ … (i)

Similarly, $V_B - V_C = I_1 Q$ and $V_D - V_C = I_2 S$ , giving $I_1 Q = I_2 S$ … (ii)

Dividing (i) by (ii): $\dfrac{P}{Q} = \dfrac{R}{S}$ . Done.

Variant 1: The Meter Bridge

The meter bridge replaces $R$ and $S$ with a uniform resistance wire of length 1 m. A sliding contact (jockey) divides the wire into two pieces of lengths $\ell_1$ and $\ell_2 = 100 - \ell_1$ cm.

Since resistance is proportional to length for a uniform wire, $R/S = \ell_1/\ell_2$ . The balance condition becomes:

$\frac{P}{Q} = \frac{\ell_1}{100 - \ell_1}$

You set $P$ to a known resistance, $Q$ to an unknown resistance, slide the jockey until the galvanometer reads zero, and read off $\ell_1$ . Then $Q = P \cdot (100 - \ell_1)/\ell_1$ .

This is how class 12 lab measurements of unknown resistance work.

Variant 2: Unbalanced Bridge

When $P/Q \neq R/S$ , current flows through the galvanometer. Calculating it requires Kirchhoff’s laws or mesh analysis — harder problem.

For the special case where the galvanometer has very high resistance ( $R_g \to \infty$ ): no current flows through it (effectively open circuit). The bridge behaves as two parallel branches: $(P + Q)$ in parallel with $(R + S)$ .

For $R_g \to 0$ (galvanometer is a short circuit): the bridge becomes ( $P$ parallel $R$ ) in series with ( $Q$ parallel $S$ ).

In between, the full Kirchhoff method is needed — this is the JEE Advanced trap question.

Variant 3: Wheatstone Hidden in a Network

Sometimes a circuit problem looks like 5 resistors in a complicated mess, but if you recognize a Wheatstone bridge with the central resistor as the “galvanometer,” you can simplify it dramatically.

If the bridge is balanced, the central resistor carries no current — you can delete it (open the connection) without changing the rest of the circuit. The remaining four resistors form two simple series-parallel paths.

This trick saves enormous time on JEE problems with 5–6 resistors. Always check for the balance ratio first.

Solved Examples

Example 1 (Easy, CBSE)

In a Wheatstone bridge, $P = 2\,\Omega$ , $Q = 3\,\Omega$ , $R = 4\,\Omega$ . Find $S$ for balance.

$P/Q = R/S \Rightarrow 2/3 = 4/S \Rightarrow S = 6\,\Omega$ .

Example 2 (Medium, NEET)

In a meter bridge, an unknown resistance $X$ is in the right gap and a $5\,\Omega$ standard is in the left gap. Balance is obtained at $\ell = 40$ cm from the left.

Balance: $5/X = 40/60 \Rightarrow X = 7.5\,\Omega$ .

Example 3 (Hard, JEE Advanced)

Five resistors $1\,\Omega, 2\,\Omega, 3\,\Omega, 6\,\Omega$ , and the central one $4\,\Omega$ form a Wheatstone-like network. Find the equivalent resistance between the battery terminals.

Check balance: $P/Q = 1/2$ and $R/S = 3/6 = 1/2$ . ✓ Balanced. The central $4\,\Omega$ carries no current. Remove it.

Now: $(1 + 2) = 3\,\Omega$ in parallel with $(3 + 6) = 9\,\Omega$ . Equivalent: $\dfrac{3 \times 9}{12} = 2.25\,\Omega$ .

Exam-Specific Tips

JEE Main weightage. Wheatstone shows up directly in 1 question every 2 shifts on average. Pure balance condition + straightforward calculation.

JEE Advanced. Expect “hidden” Wheatstone in larger circuits. Always test the balance ratio when you see 5 resistors arranged in a diamond-like shape.

NEET. Almost always meter bridge — direct application, finding unknown resistance from $\ell_1$ .

CBSE class 12 boards. A 3- or 5-marker on either the principle or a simple meter bridge calculation. Free marks if you know the derivation.

Common Mistakes to Avoid

Mistake 1: Wrong order of resistors. $P/Q = R/S$ assumes a specific layout. Mislabel $P, Q, R, S$ and the formula gives nonsense. Always draw the diamond with $P$ and $R$ on the same side as the battery’s positive terminal.

Mistake 2: Treating balanced bridge with current. When balanced, the galvanometer carries no current — but the rest of the circuit does. Don’t conclude that the entire current is zero.

Mistake 3: Forgetting to check for hidden Wheatstone. Students laboriously apply Kirchhoff to 5-resistor networks that simplify in 2 lines via Wheatstone. Always check $P/Q = R/S$ first.

Mistake 4: Mixing meter bridge ratios. $\ell_1$ is on the same side as the resistor in the formula’s numerator. Get this wrong and you get the reciprocal.

Mistake 5: Assuming galvanometer resistance affects balance. It doesn’t. At balance, no current flows through the galvanometer, so its resistance is irrelevant.

Practice Questions

Q1. In a Wheatstone bridge, $P = 5\,\Omega$ , $Q = 10\,\Omega$ , $S = 6\,\Omega$ . Find $R$ for balance.

$P/Q = R/S \Rightarrow 1/2 = R/6 \Rightarrow R = 3\,\Omega$ .

Q2. In a meter bridge, balance is at 25 cm from the left when standard resistance $S = 3\,\Omega$ is in the right gap. Find the unknown $X$ in the left gap.

$X/3 = 25/75 \Rightarrow X = 1\,\Omega$ .

Q3. A bridge has $P = Q = R = S = 2\,\Omega$ and a galvanometer of $4\,\Omega$ across the diagonal. Battery is $4$ V. Find current through the galvanometer.

Bridge is balanced ( $P/Q = R/S$ ), so galvanometer current is zero.

Q4. Five resistors of $2\,\Omega, 4\,\Omega, 6\,\Omega, 12\,\Omega$ form the four arms; a $5\,\Omega$ resistor is the central one. Find the equivalent resistance.

Check: $2/4 = 1/2$ , $6/12 = 1/2$ . Balanced. Remove $5\,\Omega$ . $(2+4) = 6$ parallel $(6+12) = 18$ : $\dfrac{6 \times 18}{24} = 4.5\,\Omega$ .

Q5. Why is a Wheatstone bridge more accurate than directly measuring resistance with an ammeter and voltmeter?

The bridge measures the null point (zero galvanometer current) rather than absolute current/voltage values. Null measurements are unaffected by instrument imperfections, so they are more accurate.

Q6. In a meter bridge, on interchanging the resistors in the two gaps, the balance shifts by 20 cm. If the original balance was at 30 cm, find the ratio of the resistances.

Original: $X/Y = 30/70 = 3/7$ . After interchange: balance at 50 cm, so $Y/X = 50/50 = 1$ — but that means $X = Y$ , contradiction. Re-check: balance shifts to $30 + 20 = 50$ cm. Then $Y/X = 50/50 = 1 \Rightarrow X = Y$ , but original ratio was $3/7$ . The shift cannot be 20 cm if $X \neq Y$ . Likely problem intends shift to 50 cm from interchange of right side; ratio is 3:7.

Q7. A Wheatstone bridge with all arms $5\,\Omega$ and a battery of EMF $5$ V (no internal resistance) is balanced. Find the total current drawn from the battery.

Balanced bridge — galvanometer has zero current. Two parallel branches each of resistance $10\,\Omega$ . Equivalent $5\,\Omega$ . Current $= 5/5 = 1$ A.

Q8. Why does the meter bridge use a uniform wire instead of a wound resistance coil?

A uniform wire ensures resistance is exactly proportional to length, so the balance condition reduces to a length ratio. Coils would have non-uniform contributions and be harder to calibrate by length.

FAQs

Q: Why doesn’t the EMF of the battery appear in the balance condition?

The balance condition is a ratio of resistances, derived by setting the galvanometer current to zero. The battery EMF only affects the magnitude of currents in the arms, not the ratio. So the balance point doesn’t depend on EMF.

Q: Can the bridge work with AC?

Yes — but for AC, you replace resistors with impedances. The condition becomes $Z_1/Z_2 = Z_3/Z_4$ , requiring matching of both magnitude and phase. This is the AC bridge (Wien, Maxwell, Anderson) — beyond JEE syllabus but appears in JEE Advanced occasionally.

Q: Why is the meter bridge sensitive when balance is near the middle?

Sensitivity (change in galvanometer current per change in length) is highest when $\ell_1 \approx 50$ cm. So you should choose $P$ such that the balance falls between 30 and 70 cm.

Q: Can the bridge be used to measure capacitance?

Yes — using AC and replacing resistors with capacitors and inductors. The DC version measures only resistance.

Q: What if the galvanometer always shows current — can the bridge ever balance?

If the four resistors are in some ratio that can’t satisfy $P/Q = R/S$ , balance is impossible with that combination. You’d need to swap a resistor.

Q: Is the Wheatstone bridge the same as the slide wire potentiometer?

No. Both are null-detection instruments, but the potentiometer measures EMF/voltage, while the Wheatstone bridge measures resistance. Different geometry, different purpose.

Q: How accurate is a meter bridge in practice?

Class 12 lab meter bridges typically give 1–2% accuracy. Sources of error include non-uniformity of the wire, contact resistance at the jockey, and the eye’s ability to detect zero galvanometer deflection.