Step 1: Analyzing the Initial State (Switch Open)

When the switch is open, the circuit consists of two parallel branches. Each branch contains two resistors in series. The ammeter reads a total current $I = 2.5 \, \text{A}$.

We are given that in this state, two resistors dissipate $50 \, \text{W}$ each, and the other two dissipate $200 \, \text{W}$ each. Since the resistors in a single branch share the same current, and $P = I^2 R$, the ratio of powers equals the ratio of resistances.

$$ \frac{P_2}{P_1} = \frac{200}{50} = 4 \implies R_{\text{high}} = 4 R_{\text{low}} $$

Assuming the branches are symmetric (standard for such bridge problems), each branch has one “low” resistor ($x$) and one “high” resistor ($4x$). To ensure the bridge is unbalanced (so the switch makes a difference), we assign them in a crossed configuration:

• Top Branch: $x$ followed by $4x$.
• Bottom Branch: $4x$ followed by $x$.

Since the branches are identical in total resistance ($5x$), the total current $2.5 \, \text{A}$ splits equally. Current in each branch is $i = 1.25 \, \text{A}$.

Using the power rating for the smaller resistor:

$$ i^2 x = 50 \implies (1.25)^2 x = 50 $$ $$ 1.5625 x = 50 \implies x = 32 \, \Omega $$

So, the resistances are:

$$ R_{\text{low}} = 32 \, \Omega $$ $$ R_{\text{high}} = 4 \times 32 = 128 \, \Omega $$

Step 2: Analyzing the Final State (Switch Closed)

When the switch is closed, it connects the midpoints of the two branches. The circuit topology changes to two parallel blocks in series.

• Block 1: Top-left ($32 \, \Omega$) in parallel with Bottom-left ($128 \, \Omega$).
• Block 2: Top-right ($128 \, \Omega$) in parallel with Bottom-right ($32 \, \Omega$).

We are told the voltage is adjusted so the total current is again $I = 2.5 \, \text{A}$.

Let’s calculate the current split in Block 1. The total current $2.5 \, \text{A}$ enters the junction of $32 \, \Omega$ and $128 \, \Omega$. By the current divider rule:

Current through top-left ($32 \, \Omega$):

$$ I_{32} = I_{\text{total}} \times \frac{128}{32 + 128} = 2.5 \times \frac{128}{160} = 2.5 \times 0.8 = 2.0 \, \text{A} $$

Current through bottom-left ($128 \, \Omega$):

$$ I_{128} = I_{\text{total}} \times \frac{32}{32 + 128} = 2.5 \times \frac{32}{160} = 2.5 \times 0.2 = 0.5 \, \text{A} $$

Step 3: Calculating Power Dissipation

Now we calculate the power for these currents.

Power in $32 \, \Omega$ resistor:

$$ P_{32} = I_{32}^2 \times 32 = (2.0)^2 \times 32 = 4 \times 32 = 128 \, \text{W} $$

Power in $128 \, \Omega$ resistor:

$$ P_{128} = I_{128}^2 \times 128 = (0.5)^2 \times 128 = 0.25 \times 128 = 32 \, \text{W} $$

Due to symmetry, Block 2 yields the same values.