A series LCR circuit is designed to resonate at an angular frequency `omega_(0)=10^5 ` rad/s. The circuit draws 16 W power from 120 V source at resonance. The value of resistance 'R' in the circuit is ______ `Omega`.

Answer

To find the value of resistance \( R \) in the given series LCR circuit, we can follow these steps:

### Step 1: Understand the power formula in an LCR circuit at resonance
At resonance, the impedance \( Z \) of the circuit is equal to the resistance \( R \). The power \( P \) consumed in the circuit is given by the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the voltage across the circuit.

### Step 2: Substitute the known values into the power formula
We know from the problem statement:
- Power \( P = 16 \, \text{W} \)
- Voltage \( V = 120 \, \text{V} \)

Substituting these values into the power formula gives:
\[ 16 = \frac{120^2}{R} \]

### Step 3: Rearrange the equation to solve for \( R \)
Rearranging the equation to isolate \( R \):
\[ R = \frac{120^2}{16} \]

### Step 4: Calculate \( R \)
Now, calculate \( 120^2 \):
\[ 120^2 = 14400 \]
Now substitute this value back into the equation for \( R \):
\[ R = \frac{14400}{16} \]
Calculating this gives:
\[ R = 900 \, \Omega \]

### Final Answer
Thus, the value of resistance \( R \) in the circuit is:
\[ \boxed{900 \, \Omega} \]

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