Step 1: Circuit Analysis and Variables

Let the resistance of the bulb be $r$. The circuit is supplied by a voltage $V = 54 \, \text{V}$. We denote the potential at the positive terminal (left) as $54 \, \text{V}$ and the negative terminal (right) as $0 \, \text{V}$.

Let the potential at the junction above the bulb be $V_C$ and the potential at the junction below the bulb be $V_D$.

Step 2: Case 1 – Switch Closed

When the switch is closed, the point $C$ is directly connected to the zero potential terminal (since the switch has negligible resistance). Therefore, $V_C = 0 \, \text{V}$.

The bulb is connected between $D$ and $C$ (Ground). The resistor of $90 \, \Omega$ (bottom right) is also connected between $D$ and Ground. Thus, the bulb and the bottom-right resistor are in parallel.

The equivalent resistance of this parallel combination is:

$$ R_{\parallel} = \frac{90 r}{90 + r} $$

This combination is in series with the $180 \, \Omega$ resistor. The voltage at node $D$, which is the voltage across the bulb ($V_{\text{closed}}$), is determined by the potential divider rule:

$$ V_{\text{closed}} = 54 \times \frac{R_{\parallel}}{180 + R_{\parallel}} $$

Substituting $R_{\parallel}$:

$$ V_{\text{closed}} = 54 \times \frac{\frac{90r}{90+r}}{180 + \frac{90r}{90+r}} = 54 \times \frac{90r}{180(90+r) + 90r} $$ $$ V_{\text{closed}} = 54 \times \frac{90r}{16200 + 180r + 90r} = \frac{4860r}{16200 + 270r} $$

Dividing numerator and denominator by 270:

$$ V_{\text{closed}} = \frac{18r}{60 + r} $$

Step 3: Case 2 – Switch Open

When the switch is open, the branch containing the switch carries no current. The current flows from the source through the top $90 \, \Omega$ resistor, then through the bulb, and enters node $D$. From the source, current also flows through the bottom $180 \, \Omega$ resistor to node $D$. The combined current then flows through the bottom-right $90 \, \Omega$ resistor to ground.

Let $i$ be the current flowing through the top branch (top 90 $\Omega$ and Bulb). The potential drops are:

$$ V_C = 54 – 90i $$ $$ V_D = V_C – ir = 54 – 90i – ir = 54 – i(90+r) $$

Apply Kirchhoff’s Current Law (KCL) at node $D$:

$$ \text{Current from bottom left} + \text{Current from Bulb} = \text{Current to bottom right} $$ $$ \frac{54 – V_D}{180} + i = \frac{V_D}{90} $$

Multiplying the entire equation by 180:

$$ (54 – V_D) + 180i = 2V_D $$ $$ 54 + 180i = 3V_D \implies V_D = 18 + 60i $$

Now, equate the two expressions for $V_D$:

$$ 18 + 60i = 54 – i(90+r) $$ $$ i(60 + 90 + r) = 54 – 18 $$ $$ i(150 + r) = 36 \implies i = \frac{36}{150 + r} $$

The voltage across the bulb in the open case is $V_{\text{open}} = i \times r$:

$$ V_{\text{open}} = \frac{36r}{150 + r} $$

Step 4: Solving for Resistance and Voltage

The problem states the bulb glows equally bright, meaning the voltage drop across it is the same in both cases:

$$ V_{\text{closed}} = V_{\text{open}} $$ $$ \frac{18r}{60 + r} = \frac{36r}{150 + r} $$

Dividing both sides by $18r$:

$$ \frac{1}{60 + r} = \frac{2}{150 + r} $$ $$ 150 + r = 2(60 + r) $$ $$ 150 + r = 120 + 2r $$ $$ r = 30 \, \Omega $$

Finally, calculate the voltage drop:

$$ V = \frac{18(30)}{60 + 30} = \frac{540}{90} = 6 \, \text{V} $$