The following table gives the reactance and rms voltage across the elements of a series RCL circuit :
`{:("Circuit element","Reactance","Voltage across element"),("Resistor", 2.00xx10^2 Omega, 86V),("Inductor",3.77xx10^2 Omega,162V):}`
What is the power factor for this circuit ?

Options

A. `0.40`
B. `0.81`
C. `0.47`
D. `1.4`

Answer

To find the power factor for the given series RCL circuit, we will follow these steps:

### Step 1: Identify the given values
We have the following values from the table:
- Voltage across the resistor (Vr) = 86 V
- Voltage across the inductor (Vl) = 162 V

### Step 2: Calculate the total voltage (Vs)
The total voltage in a series RCL circuit can be calculated using the Pythagorean theorem, since the voltages across the resistor and inductor are perpendicular to each other.

\[ Vs = \sqrt{Vr^2 + Vl^2} \]

Substituting the values:

\[ Vs = \sqrt{(86)^2 + (162)^2} \]

Calculating \(86^2\) and \(162^2\):

\[ 86^2 = 7396 \]
\[ 162^2 = 26244 \]

Now, adding these values:

\[ Vs = \sqrt{7396 + 26244} = \sqrt{33640} \]

Calculating the square root:

\[ Vs \approx 183.412 \, V \]

### Step 3: Calculate the power factor (cos φ)
The power factor is defined as the ratio of the voltage across the resistor to the total voltage:

\[ \text{Power Factor} = \cos \phi = \frac{Vr}{Vs} \]

Substituting the values we have:

\[ \text{Power Factor} = \frac{86}{183.412} \]

Calculating this value:

\[ \text{Power Factor} \approx 0.468 \]

Rounding to two decimal places gives us:

\[ \text{Power Factor} \approx 0.47 \]

### Final Answer
The power factor for the circuit is approximately **0.47**.

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