A condenser of capacity 1 `mu` F and a resistance 0.5 mega-ohm are connected in series with a DC supply of 2 V. The time constant of circuit is

Options

A. 2 s
B. 1 s
C. 0.5 s
D. 0.25 s

Answer

To find the time constant of the circuit, we can use the formula for the time constant (τ) in an RC (resistor-capacitor) circuit, which is given by:

\[ \tau = R \times C \]

Where:
- \( \tau \) is the time constant in seconds,
- \( R \) is the resistance in ohms,
- \( C \) is the capacitance in farads.

### Step 1: Identify the values of resistance and capacitance
- The resistance \( R \) is given as \( 0.5 \) mega-ohms.
- We need to convert this to ohms:
\[ R = 0.5 \, \text{mega-ohms} = 0.5 \times 10^6 \, \text{ohms} = 500000 \, \text{ohms} \]

- The capacitance \( C \) is given as \( 1 \, \mu F \).
- We need to convert this to farads:
\[ C = 1 \, \mu F = 1 \times 10^{-6} \, \text{farads} \]

### Step 2: Substitute the values into the time constant formula
Now, we can substitute the values of \( R \) and \( C \) into the time constant formula:
\[ \tau = R \times C = (500000 \, \text{ohms}) \times (1 \times 10^{-6} \, \text{farads}) \]

### Step 3: Calculate the time constant
Now, we perform the multiplication:
\[ \tau = 500000 \times 1 \times 10^{-6} = 0.5 \, \text{seconds} \]

### Conclusion
The time constant of the circuit is \( 0.5 \) seconds.

### Final Answer
\[ \text{The time constant } \tau = 0.5 \, \text{seconds} \]

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