Obtain the condition for bridge balance in Wheatstone's bridge.

Question 66 of 102

Q. Obtain the condition for bridge balance in Wheatstone's bridge.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2017Subjective· 5mImportance★★★★★

65% · 66/102 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Wheatstone Bridge

The Wheatstone Bridge – From Intuition to Precision

Imagine you have a single unknown resistor and you want to find its value. You could use an ohmmeter, but those are not always accurate for very small or very large resistances. A more elegant method is to compare it against known resistances in a circuit that acts like a balance scale — that is the Wheatstone bridge.

The core idea is simple: make the voltage at two points equal, so no current flows between them. When that happens, you know the ratio of the resistances.

The Circuit Layout

The bridge has four resistors arranged in a diamond shape:

        A
       / \
      P   Q
     /     \
    B-------C
     \     /
      R   S
       \ /
        D

A battery is connected across A and D. A sensitive galvanometer (G) is connected between B and C. The four resistors are labelled P, Q, R, and S. Usually, three are known and one (say, S) is unknown.

The Intuition: Two Voltage Dividers

Look at the left side: from A to D through P and R. That is a voltage divider. The voltage at B is a fraction of the battery voltage, determined by the ratio of P to R.

Now look at the right side: from A to D through Q and S. That is another voltage divider. The voltage at C is a fraction of the battery voltage, determined by the ratio of Q to S.

If the voltage at B equals the voltage at C, then no current flows through the galvanometer — the bridge is balanced.

The Condition for Balance

When the bridge is balanced, the voltage drop across P equals the voltage drop across Q (since both start at A), and the voltage drop across R equals the voltage drop across S (since both end at D). From the voltage divider rule:

Voltage at B: $V_B = V_A \cdot \frac{R}{P+R}$
Voltage at C: $V_C = V_A \cdot \frac{S}{Q+S}$

Setting $V_B = V_C$ gives:

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

Cross-multiply:

R(Q+S) = S(P+R)

RQ + RS = SP + SR

The $RS$ terms cancel, leaving:

RQ = SP

Or, rearranged:

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

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

That is the balance condition of the Wheatstone bridge. When this holds, the galvanometer shows zero deflection.

Measuring an Unknown Resistance

Suppose S is unknown. You set P, Q, and R to known values. You adjust R (or the ratio P/Q) until the galvanometer reads zero. Then you compute:

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

This is why the bridge is so useful: you do not need to measure current or voltage accurately — you only need to detect when current is zero. That is far more sensitive and precise.

Tip

In practice, P and Q are often made equal (a 1:1 ratio), so the unknown S simply equals R. This is the "equal-arm" bridge.

Why It Works So Well

The galvanometer is a null detector — it only tells you whether current is flowing, not how much. This eliminates errors from meter calibration, battery voltage fluctuations, and temperature effects. The accuracy depends only on the precision of the known resistors. …

Start your 14-day free trial to unlock the full solution →

Applying Kirchhoff's laws to a Wheatstone's bridge under the balanced (zero galvanometer current) condition shows that the ratios of opposite arms must be equal, $P/Q=R/S$ .

The bridge circuit

A Wheatstone's bridge consists of four resistances $P,\,Q,\,R,\,S$ connected to form a closed quadrilateral $ABCD$ : $P$ between $A$ and $B$ , $Q$ between $B$ and $C$ , $S$ between $A$ and $D$ , and $R$ between $D$ and $C$ . A battery (with a key) is connected across the diagonal $AC$ , and a galvanometer (with a key) is connected across the other diagonal $BD$ .

Balanced condition

The bridge is said to be balanced when the galvanometer shows zero deflection, i.e. no current flows through it ( $I_g=0$ ). This means points $B$ and $D$ are at the same potential.

Applying Kirchhoff's current law (junction rule)

Since no current flows through the galvanometer branch $BD$ , the entire current entering junction $B$ from $A$ (through $P$ ) must continue on to $C$ (through $Q$ ); call this current $I_1$ . Similarly, the current entering $D$ from $A$ (through $S$ ) continues on to $C$ (through $R$ ); call this current $I_2$ .

Applying Kirchhoff's voltage law (loop rule)

Because $B$ and $D$ are at the same potential, the potential drop from $A$ to $B$ (through $P$ ) must equal the potential drop from $A$ to $D$ (through $S$ ):

$I_1P=I_2S \qquad \text{...(1)}$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ7 — A galvanometer of resistance 100 Ω which can measure a maximum current…→Next questionQ9 — How can e.m.f. of two cells be compared using potentiometer ?

Q.Obtain the condition for bridge balance in Wheatstone's bridge.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days