Instantaneous voltage and instantaneoss current in an L-R circuit in AC is V=100sin (100)t and `i=10sin(100t-pi//4)`. Match the following table,
`{:(,(A),R,(P),(1)/(10sqrt2)"SI units"),(,(B),X_(L),(Q),5sqrt2 "SI unit"),(,(C),L,(R),10sqrt2 "SI units"),(,(D),"Average power in one cycle",(S),"None"):}`

Answer

To solve the problem, we need to analyze the given instantaneous voltage and current in the L-R circuit and match them with the appropriate values in the table provided.

### Step 1: Identify the given parameters
The instantaneous voltage and current are given as:
- Voltage: \( V(t) = 100 \sin(100t) \)
- Current: \( I(t) = 10 \sin(100t - \frac{\pi}{4}) \)

From this, we can extract:
- Peak voltage, \( V_0 = 100 \, \text{V} \)
- Peak current, \( I_0 = 10 \, \text{A} \)
- Phase difference, \( \phi = \frac{\pi}{4} \)

### Step 2: Calculate the impedance (Z)
Using the relationship between peak voltage and peak current:
\[ Z = \frac{V_0}{I_0} = \frac{100}{10} = 10 \, \Omega \]

### Step 3: Calculate the resistance (R)
Using the cosine of the phase angle:
\[ \cos(\phi) = \frac{R}{Z} \]
Substituting \( \phi = \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{R}{10} \implies \frac{1}{\sqrt{2}} = \frac{R}{10} \]
Thus,
\[ R = 10 \cdot \frac{1}{\sqrt{2}} = 5\sqrt{2} \, \Omega \]

### Step 4: Calculate the inductive reactance (X_L)
Using the impedance formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the known values:
\[ 10 = \sqrt{(5\sqrt{2})^2 + X_L^2} \]
Calculating \( (5\sqrt{2})^2 = 50 \):
\[ 10 = \sqrt{50 + X_L^2} \]
Squaring both sides:
\[ 100 = 50 + X_L^2 \implies X_L^2 = 50 \implies X_L = 5\sqrt{2} \, \Omega \]

### Step 5: Calculate the self-inductance (L)
Using the relationship between inductive reactance and self-inductance:
\[ X_L = \omega L \]
Where \( \omega = 100 \, \text{rad/s} \):
\[ 5\sqrt{2} = 100L \implies L = \frac{5\sqrt{2}}{100} = \frac{\sqrt{2}}{20} \, \text{H} \]

### Step 6: Calculate the average power (P)
The average power in an AC circuit is given by:
\[ P = I_{\text{rms}} V_{\text{rms}} \cos(\phi) \]
Where:
- \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \)
- \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} \)

Calculating \( P \):
\[ P = \left(\frac{10}{\sqrt{2}}\right) \left(\frac{100}{\sqrt{2}}\right) \cos\left(\frac{\pi}{4}\right) \]
\[ = \left(\frac{1000}{2}\right) \cdot \frac{1}{\sqrt{2}} = 500\sqrt{2} \, \text{W} \]

### Summary of Results
- \( R = 5\sqrt{2} \, \Omega \) (matches with Q)
- \( X_L = 5\sqrt{2} \, \Omega \) (matches with Q)
- \( L = \frac{\sqrt{2}}{20} \, \text{H} \) (matches with P)
- Average power \( P = 500\sqrt{2} \, \text{W} \) (matches with S)

### Final Matching
- A matches with Q
- B matches with Q
- C matches with P
- D matches with S (None)