Comparison of Power Dissipation in Three Resistor Configurations

To determine the correct relationship between the powers dissipated by the three configurations, we find the equivalent resistance $R_{\text{eq}}$ of each circuit connected across the $V = 3\text{ V}$ battery. The power dissipated in each case is given by:
$P = \frac{V^2}{R_{\text{eq}}}$

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1. Configuration $R_2$:

• The circuit is a symmetric Wheatstone bridge consisting of five $1\,\Omega$ resistors.
• The four outer arms each have resistance $R = 1\,\Omega$, and the central resistor has resistance $R = 1\,\Omega$.
• The ratio of the resistances of opposite arms is balanced:
  $\frac{R_{\text{top-left}}}{R_{\text{bottom-left}}} = \frac{1\,\Omega}{1\,\Omega} = \frac{R_{\text{top-right}}}{R_{\text{bottom-right}}}$
• Therefore, the potential difference across the central $1\,\Omega$ resistor is zero, and no current flows through it.
• The remaining four resistors form two parallel branches, each having two $1\,\Omega$ resistors in series:
  $R_{\text{branch}} = 1\,\Omega + 1\,\Omega = 2\,\Omega$
• The equivalent resistance of configuration $R_2$ is:
  $R_{\text{eq}, 2} = \frac{2\,\Omega \times 2\,\Omega}{2\,\Omega + 2\,\Omega} = 1\,\Omega$
• The power dissipated in configuration $R_2$ is:
  $P_2 = \frac{V^2}{R_{\text{eq}, 2}} = \frac{3^2}{1} = 9\text{ W}$

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2. Configuration $R_3$:

• Configuration $R_3$ consists of a four-resistor bridge network connected in series with an additional $1\,\Omega$ resistor at the bottom.
• The upper diamond bridge consists of two parallel branches, each with two $1\,\Omega$ resistors in series:
  $R_{\text{bridge}} = \frac{(1\,\Omega + 1\,\Omega) \times (1\,\Omega + 1\,\Omega)}{(1\,\Omega + 1\,\Omega) + (1\,\Omega + 1\,\Omega)} = \frac{2 \times 2}{2 + 2} = 1\,\Omega$
• This bridge is in series with the bottom $1\,\Omega$ resistor:
  $R_{\text{eq}, 3} = R_{\text{bridge}} + 1\,\Omega = 1\,\Omega + 1\,\Omega = 2\,\Omega$
• The power dissipated in configuration $R_3$ is:
  $P_3 = \frac{V^2}{R_{\text{eq}, 3}} = \frac{3^2}{2} = 4.5\text{ W}$

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3. Configuration $R_1$:

• By analyzing the node connections of configuration $R_1$:
  • The parallel and series combinations of the five $1\,\Omega$ resistors reduce to an equivalent resistance of:
    $R_{\text{eq}, 1} = 1.5\,\Omega$
• The power dissipated in configuration $R_1$ is:
  $P_1 = \frac{V^2}{R_{\text{eq}, 1}} = \frac{3^2}{1.5} = 6\text{ W}$

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Comparison:

Comparing the power dissipated in the three configurations:
$P_2 = 9\text{ W}, \quad P_1 = 6\text{ W}, \quad P_3 = 4.5\text{ W}$

Thus:
$\text{P2} > \text{P1} > \text{P3}$

Therefore, the correct option is (C).