Q.State Kirchhoff's laws of electrical network.

The assertion is true because the Wheatstone bridge is a reciprocal network, but the reason given (a trivial algebraic manipulation) does not explain why interchanging the cell and galvanometer preserves balance. The correct option is (B).

(d) Both the Statements are true …

A Wheatstone bridge is balanced when P/Q = R/S — only the four arm resistances matter. Changing an arm's resistance upsets balance; interchanging the cell and galvanometer, or changing their values/emf, does not. Option (a).

Step 1 — balance condition (NCERT/CBSE Class 12 Physics, Current Electricity): P/Q = R/S for the four ratio arms.

Step 2 — evaluate each option:

• (a) Changing arm resistances alters the P/Q or R/S ratio → balance changes. This is the correct choice. …

A Wheatstone bridge is balanced when the ratio of resistances in the two arms is equal on both sides, so the galvanometer carries no current.

A Wheatstone bridge has four resistances P, Q, R, S arranged in a diamond, with a galvanometer connected across the middle (between the P–Q junction and the R–S junction) and a battery driving current through the outer loop. The bridge is said to be balanced when the potential at the galvanometer's two terminals is equal, so no current flows through it ($I_g = 0$).

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A Wheatstone bridge is a four-arm resistance network used to accurately measure an unknown resistance by balancing it against three known resistances.

The Wheatstone bridge consists of four resistors arranged in a diamond/quadrilateral, with a galvanometer connected across one diagonal and a battery across the other. By adjusting the known resistances until the galvanometer shows zero deflection (balanced condition, $P/Q = R/S$), the unknown resistance can be calculated precisely from the …

The Wheatstone-bridge balance condition AB/BC = AD/DC is satisfied exactly, so no current flows through the galvanometer.

For the bridge A-B-C-D with the galvanometer across the B-D diagonal, the bridge is balanced when

$\dfrac{AB}{BC}=\dfrac{AD}{DC}$

Here $AB=6\,\Omega,\ BC=3\,\Omega,\ AD=12\,\Omega,\ DC=6\,\Omega$:

$\dfrac{6}{3}=2,\qquad \dfrac{12}{6}=2$ …

In a balanced Wheatstone bridge, the ratio of the two arms on one side equals the ratio on the other side: P/Q = R/S.

Label the bridge arms as given: AB = P = 30 Ω, BC = Q = 10 Ω, AD = R = 18 Ω, DC = S (unknown), with the galvanometer connected between B and D.

For a balanced Wheatstone bridge (no current through the galvanometer), the balance condition is:

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The Wheatstone bridge principle: four resistances arranged in a closed network are "balanced" (zero galvanometer current) exactly when the ratio of resistances in one pair of adjacent arms equals that of the other pair.

A Wheatstone bridge consists of four resistances $P,Q,R,S$ forming a closed quadrilateral (arms $AB=P$, $BC=Q$, $AD=R$, $DC=S$), with a battery connected across one diagonal ($A$–$C$) and a galvanometer across the other diagonal ($B$–$D$).

The bridge is said to be balanced when the galvanometer shows no deflection, i.e. no current flows through the $BD$ arm — this happens when points $B$ and $D$ are at the same potential.

Under this balanced condition, applying Kirchhoff's laws to the two loops gives the principle:

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Balanced bridge → galvanometer carries no current; the network reduces to two 2R branches in parallel = R: option (a).

In a balanced Wheatstone bridge, the bridge (galvanometer) arm carries no current, because the potentials at its two ends are equal. Therefore the galvanometer branch can be removed without changing anything.

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