A resistance of `40` ohms and an inductance of `95.5` millihenry are connected in series in a `50` cycle/sec AC circuit. The impedence of this combination is very nearly

Options

A. 30 ohm
B. 40 ohm
C. 50 ohm
D. 60 ohm

Answer

To find the impedance of a circuit consisting of a resistance and an inductance connected in series in an AC circuit, we can follow these steps:

### Step 1: Identify the Given Values
- Resistance, \( R = 40 \, \Omega \)
- Inductance, \( L = 95.5 \, \text{mH} = 95.5 \times 10^{-3} \, \text{H} \)
- Frequency, \( f = 50 \, \text{Hz} \)

### Step 2: Calculate Angular Frequency
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2 \pi f \]
Substituting the given frequency:
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]

### Step 3: Calculate Inductive Reactance
The inductive reactance \( X_L \) is calculated using the formula:
\[ X_L = \omega L \]
Substituting the values of \( \omega \) and \( L \):
\[ X_L = 100 \pi \times 95.5 \times 10^{-3} \]
Calculating this gives:
\[ X_L \approx 100 \times 3.14 \times 0.0955 \approx 30 \, \Omega \quad (\text{using } \pi \approx 3.14) \]

### Step 4: Calculate Impedance
The total impedance \( Z \) in a series circuit with resistance and inductive reactance is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values of \( R \) and \( X_L \):
\[ Z = \sqrt{40^2 + (30)^2} \]
Calculating:
\[ Z = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \Omega \]

### Final Answer
The impedance of the combination is very nearly \( 50 \, \Omega \).

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