An electric motor which is loaded has an effective resistance of `30Omega` and an inductive reactance of `40Omega` . If the motor is powered by a source with a maximum voltage of 420 V, the maximum current is

Options

A. `6A`
B. `8.4 A`
C. `10A`
D. `12A`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify Given Values
- Effective resistance (R) = 30 Ω
- Inductive reactance (X_L) = 40 Ω
- Maximum voltage (V_max) = 420 V

### Step 2: Calculate Impedance (Z)
The impedance (Z) in an R-L circuit is calculated using the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]
Substituting the given values:
\[ Z = \sqrt{(30)^2 + (40)^2} \]
Calculating the squares:
\[ Z = \sqrt{900 + 1600} = \sqrt{2500} \]
Taking the square root:
\[ Z = 50 \, \Omega \]

### Step 3: Calculate Maximum Current (I_max)
Using Ohm's Law, the maximum current (I_max) can be calculated using the formula:
\[ I_{max} = \frac{V_{max}}{Z} \]
Substituting the known values:
\[ I_{max} = \frac{420 \, V}{50 \, \Omega} \]
Calculating the current:
\[ I_{max} = 8.4 \, A \]

### Final Answer
The maximum current is:
\[ \boxed{8.4 \, A} \]

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