To solve the problem step by step, let's analyze the given information and derive the required values systematically.
### Given:
- Tube light rating: \( V_r = 60 \, V \), \( P_r = 60 \, W \)
- AC source voltage: \( V = 100 \, V \)
- Frequency: \( f = 50 \, Hz \)
### Step 1: Calculate the current through the tube light
Using the formula for power:
\[
P_r = V_r \times I
\]
We can rearrange this to find the current \( I \):
\[
I = \frac{P_r}{V_r}
\]
Substituting the given values:
\[
I = \frac{60 \, W}{60 \, V} = 1 \, A
\]
### Step 2: Calculate the voltage across the inductor
Using the relationship between the total voltage, the voltage across the resistor, and the voltage across the inductor:
\[
V = V_r + V_L
\]
Where \( V_L \) is the voltage across the inductor. Rearranging gives:
\[
V_L = V - V_r
\]
Substituting the values:
\[
V_L = 100 \, V - 60 \, V = 40 \, V
\]
### Step 3: Calculate the inductive reactance \( X_L \)
Using Ohm's law for AC circuits:
\[
V_L = I \times X_L
\]
Rearranging gives:
\[
X_L = \frac{V_L}{I}
\]
Substituting the known values:
\[
X_L = \frac{40 \, V}{1 \, A} = 40 \, \Omega
\]
### Step 4: Calculate the inductance \( L \)
The inductive reactance is related to the inductance and frequency by the formula:
\[
X_L = 2 \pi f L
\]
Rearranging to find \( L \):
\[
L = \frac{X_L}{2 \pi f}
\]
Substituting the values:
\[
L = \frac{40 \, \Omega}{2 \pi \times 50 \, Hz} = \frac{40}{100 \pi} = \frac{4}{5 \pi} \, H
\]
### Step 5: Calculate the resistance \( R \)
Using Ohm's law for the resistor:
\[
R = \frac{V_r}{I}
\]
Substituting the values:
\[
R = \frac{40 \, V}{1 \, A} = 40 \, \Omega
\]
### Conclusion
The values we have calculated are:
- Inductance \( L = \frac{4}{5 \pi} \, H \)
- Resistance \( R = 40 \, \Omega \)
### Final Answer
The correct options based on the calculations are:
- Inductance \( L = \frac{4}{5 \pi} \, H \) (Option B)
- Resistance \( R = 40 \, \Omega \) (Option D)
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