To solve the problem step by step, we will follow these steps:
### Step 1: Determine the Inductive Reactance (XL) of the Choke
Given that the choke takes a current of 10 A at 125 V and 50 Hz, we can use Ohm's law for AC circuits:
\[ V = I \cdot X_L \]
Where:
- \( V = 125 \, V \)
- \( I = 10 \, A \)
Rearranging the equation to find \( X_L \):
\[ X_L = \frac{V}{I} = \frac{125}{10} = 12.5 \, \Omega \]
### Step 2: Calculate the Resistance (R) of the Resistor
For the pure resistor, we know it takes a current of 12.5 A at the same voltage of 125 V:
\[ V = I \cdot R \]
Where:
- \( V = 125 \, V \)
- \( I = 12.5 \, A \)
Rearranging the equation to find \( R \):
\[ R = \frac{V}{I} = \frac{125}{12.5} = 10 \, \Omega \]
### Step 3: Calculate the Inductive Reactance (XL) at 40 Hz
Now, we need to find the inductive reactance when the frequency is 40 Hz. The formula for inductive reactance is:
\[ X_L = 2 \pi f L \]
We already found the inductance \( L \) from the previous calculations. We can find \( L \) using the initial frequency of 50 Hz:
Using \( X_L = 12.5 \, \Omega \):
\[ 12.5 = 2 \pi (50) L \]
Solving for \( L \):
\[ L = \frac{12.5}{2 \pi (50)} = \frac{12.5}{100 \pi} = \frac{0.125}{\pi} \, H \]
Now, substituting \( L \) back into the equation for \( X_L \) at 40 Hz:
\[ X_L = 2 \pi (40) \left(\frac{0.125}{\pi}\right) \]
Simplifying:
\[ X_L = 2 \cdot 40 \cdot 0.125 = 10 \, \Omega \]
### Step 4: Calculate the Total Impedance (Z) of the Series Circuit
The total impedance \( Z \) in an LR circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the values of \( R \) and \( X_L \):
\[ Z = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \Omega \]
### Step 5: Calculate the Current (I) in the Series Circuit
Now, we can find the current when the circuit is connected to a 100 V supply:
Using Ohm's law:
\[ I = \frac{V}{Z} \]
Where:
- \( V = 100 \, V \)
- \( Z = 10\sqrt{2} \, \Omega \)
Substituting the values:
\[ I = \frac{100}{10\sqrt{2}} = \frac{10}{\sqrt{2}} \, A \]
### Final Answer
The current in the series combination of the resistor and inductor is:
\[ I = \frac{10}{\sqrt{2}} \, A \]
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