In a series A.C. circuit , `R=100Omega`, `X_(L)=300Omega` and `X_(C )=200Omega`. The phase difference between the applied e.m.f. and the current will be

Options

A. `0`
B. `37^(@)`
C. `45^(@)`
D. `90^(@)`

Answer

To find the phase difference between the applied e.m.f. and the current in a series A.C. circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance (R) = 100 Ω
- Inductive Reactance (X_L) = 300 Ω
- Capacitive Reactance (X_C) = 200 Ω

2. **Calculate the Net Reactance (X):**
The net reactance (X) in the circuit is given by:
\[ X = X_L - X_C \]
Substituting the values:
\[ X = 300 \, \Omega - 200 \, \Omega = 100 \, \Omega \]

3. **Calculate the Impedance (Z):**
The total impedance (Z) in a series A.C. circuit is calculated using the formula:
\[ Z = \sqrt{R^2 + X^2} \]
Substituting the values of R and X:
\[ Z = \sqrt{(100 \, \Omega)^2 + (100 \, \Omega)^2} = \sqrt{10000 + 10000} = \sqrt{20000} = 100\sqrt{2} \, \Omega \]

4. **Calculate the Cosine of the Phase Difference (cos φ):**
The cosine of the phase difference (φ) is given by:
\[ \cos φ = \frac{R}{Z} \]
Substituting the values of R and Z:
\[ \cos φ = \frac{100 \, \Omega}{100\sqrt{2} \, \Omega} = \frac{1}{\sqrt{2}} \]

5. **Determine the Phase Difference (φ):**
To find the phase difference, we take the inverse cosine:
\[ φ = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
This corresponds to:
\[ φ = 45^\circ \]

### Final Answer:
The phase difference between the applied e.m.f. and the current is \( 45^\circ \).

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