Wheatstone Bridge

The ratio of resistors is $\frac{R_{AB}}{R_{BC}} = \frac{5}{2.5} = 2$ and $\frac{R_{FE}}{R_{ED}} = \frac{3}{1.5} = 2$. Since the bridge is balanced, no current flows through the central $6\,\Omega$ resistor. Equivalent resistance of the bridge network between B and E: top branch is $5 + 2.5 = 7.5\,\Omega$... In the given diagram labeling, top branch $R_{top} = 5 + 3 = 8\,\Omega$ or $(5 + 2.5)$ and bottom $(3 + 1.5)$. Specifically, parallel branches of $4\,\Omega$ and $8\,\Omega$ give $R_{BE} = \frac{4 \times 8}{4 + 8} = \frac{8}{3}\,\Omega$. Total circuit resistance is $R_{total} = \frac{8}{3} + 1.5 + 5.5 + \frac{1}{3} = \frac{9}{3} + 7 = 3 + 7 = 10\,\Omega$. The battery current is $I = \frac{V}{R_{total}} = \frac{5\text{ V}}{10\,\Omega} = 0.5\text{ A}$.

In the circuit where the diode shorts the network, no balance is possible. In the circuits where the diode D is reverse biased, the arm containing it is open, so the bridge cannot balance. In the circuit where the diode is forward biased (acting as a closed switch with its resistance $R_d$), the bridge can be balanced if $\frac{P}{Q} = \frac{R}{S}$: $\frac{10}{15} = \frac{10}{5 + R_d} \Rightarrow R_d = 10\ \Omega$ So the correct circuit is the one in which the diode is forward biased.

A Wheatstone bridge is most sensitive when all four resistances are of the same order. So for the most precise measurement of X, P and Q should be approximately equal (and small), which reduces the error in the experiment.

$\frac{P}{Q} = \frac{R}{S} = \frac{1}{3}$, so the bridge is balanced and no current flows through the galvanometer.
Resistance of the bridge $= \frac{(40)(120)}{40 + 120} = 30\ \Omega$
Current through the cell $= \frac{7}{5 + 30} = \frac{1}{5}$ A = 0.2 A

A balanced Wheatstone bridge requires $\dfrac{P}{Q} = \dfrac{R}{S'}$, where $S'$ is the resistance of the fourth arm.
$\dfrac{2}{2} = \dfrac{2}{S'}$
Therefore the effective resistance of the fourth arm should be $S' = 2\,\Omega$.
This effective resistance is the parallel combination of S and the $6\,\Omega$ resistance.
In a parallel combination: $\dfrac{1}{S'} = \dfrac{1}{6} + \dfrac{1}{S}$
$\dfrac{1}{2} = \dfrac{1}{6} + \dfrac{1}{S}$
$\dfrac{1}{S} = \dfrac{1}{2} - \dfrac{1}{6} = \dfrac{3 - 1}{6} = \dfrac{1}{3}$
$\Rightarrow S = 3\,\Omega$

The resistances $4\,\Omega$, $6\,\Omega$, $2\,\Omega$ and $3\,\Omega$ form the four arms of a Wheatstone bridge, with the middle $4\,\Omega$ resistor as the galvanometer arm.
Check the balance: $\dfrac{4}{6} = \dfrac{2}{3}$, so the bridge is balanced and no current flows through the middle $4\,\Omega$ resistor; it can be removed.
The circuit then reduces to two parallel branches: $(4 + 2) = 6\,\Omega$ and $(6 + 3) = 9\,\Omega$.
Equivalent resistance $= \dfrac{6 \times 9}{6 + 9} = \dfrac{54}{15} = \dfrac{18}{5}\,\Omega$
Current drawn from the source: $i = \dfrac{V}{18/5}$
$\Rightarrow i = \dfrac{5V}{18}$

All four arms are equal, so $\dfrac{P}{Q} = \dfrac{R}{S}$ and the bridge is balanced. No current flows through the galvanometer arm, so it can be removed.
The circuit then has two branches in parallel, each of $R + R = 2R$.
Equivalent resistance $= \dfrac{2R \times 2R}{2R + 2R} = R$
So the resistance seen by the battery, i.e. the equivalent resistance between A and B, is R.

All four arms are equal (10 $\Omega$ each), so the bridge is balanced: $\dfrac{P}{Q} = \dfrac{R}{S}$.
No current flows through the galvanometer branch, whatever its resistance (10 $\Omega$ added in series makes no difference). So that branch can be removed.
Then there are two branches of $10 + 10 = 20\ \Omega$ in parallel.
$R_{eq} = \dfrac{20 \times 20}{20 + 20} = 10\ \Omega$

The arrangement is a Wheatstone bridge with arms 3, 4 (top) and 6, 8 (bottom), and 7 $\Omega$ in the middle branch.
Check balance: $\dfrac{3}{4} = \dfrac{6}{8}$, so the bridge is balanced and no current flows through the 7 $\Omega$ resistor. It can be removed.
Top branch: $3 + 4 = 7\ \Omega$; bottom branch: $6 + 8 = 14\ \Omega$. These are in parallel.
$R = \dfrac{R_1R_2}{R_1 + R_2} = \dfrac{14 \times 7}{21} = \dfrac{14}{3}\ \Omega$