Stepwise reduction of the resistor ladder from top to bottom:

• Top apex loop: The two top resistors of $4\,\Omega$ each are in series:
  $$R_1 = 4\,\Omega + 4\,\Omega = 8\,\Omega$$
• This combination is in parallel with the upper horizontal cross-resistor of $8\,\Omega$:
  $$R_{p1} = \frac{8 \times 8}{8 + 8} = 4\,\Omega$$
• This $R_{p1} = 4\,\Omega$ is in series with the two middle-tier side resistors of $4\,\Omega$ each:
  $$R_2 = 4\,\Omega + R_{p1} + 4\,\Omega = 4 + 4 + 4 = 12\,\Omega$$
• This combination is in parallel with the second horizontal cross-resistor of $12\,\Omega$:
  $$R_{p2} = \frac{12 \times 12}{12 + 12} = 6\,\Omega$$
• Finally, this $R_{p2} = 6\,\Omega$ is in series with the two bottom-tier resistors of $6\,\Omega$ each connected to terminals A and B:
  $$R_{AB} = 6\,\Omega + R_{p2} + 6\,\Omega = 6 + 6 + 6 = 18\,\Omega$$

The condition for a balanced Wheatstone bridge is:

Given $P = 3\,\Omega, Q = 3\,\Omega, R = 3\,\Omega$:

The arm has resistance $S = 4\,\Omega$. To reduce its equivalent resistance to $S_{\text{eff}} = 3\,\Omega$, a shunt resistance $S'$ is connected in parallel with $S$:

Kirchhoff's Second Law (Voltage Law or Loop Rule) states that the algebraic sum of changes in potential around any closed loop must be zero:

This is a direct consequence of the law of conservation of energy (electrostatic force is conservative, so work done in taking a unit positive test charge around a closed circuit loop is zero).
In contrast, Kirchhoff's First Law (Junction Rule) is based on the conservation of charge.