In the circuit shown below, the switch S is closed at t =0. The magnitude of the steady state voltage, in volts, across the 6Ω resistor is ______. (round off to two decimal places).

The correct answer is 5.

Let us analyze the behavior of the circuit under steady-state conditions after the switch S is closed at $t=0$.

Step 1: Simplify the parallel combination of resistors
At the top of the circuit, the $6\Omega$ resistor and the $3\Omega$ resistor are connected in parallel. Their equivalent resistance ($R_p$) can be calculated as:
$R_p=\frac{6\times 3}{6+3}=\frac{18}{9}=2\Omega$

Step 2: Determine the state of the capacitor in DC steady state
Under DC steady-state conditions ($t\to \infty$), a capacitor acts as an open circuit because it is fully charged and does not allow any conduction current to pass through it:
$I_c=0$
Therefore, the middle branch containing the $1\mathrm{\mu F}$ capacitor and the $10\Omega$ resistor behaves as an open circuit and carries no current.

Step 3: Analyze the simplified single-loop circuit
Since no current flows through the middle branch, the circuit simplifies to a single closed loop containing:
- The $10V$ DC source
- The bottom $2\Omega$ resistor
- The equivalent parallel resistance of the top branch, $R_p=2\Omega$

The total steady-state current ($I$) flowing through this loop is:
$I=\frac{V}{R_p+2}=\frac{10}{2+2}=\frac{10}{4}=2.5A$

Step 4: Calculate the voltage across the 6Ω resistor
The voltage drop across the parallel combination of the $6\Omega$ and $3\Omega$ resistors is:
$V_p=I\times R_p=2.5A\times 2\Omega =5V$
Since the voltage across parallel branches is identical, the magnitude of the steady-state voltage across the $6\Omega$ resistor is:
$V_{6\Omega }=V_p=5V$