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 find the maximum current flowing through the electric motor, we can follow these steps:

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

### Step 2: Calculate the impedance (Z)
The impedance (Z) in an AC circuit with resistance (R) and inductive reactance (X_L) is given by the formula:
\[ Z = \sqrt{R^2 + X_L^2} \]

Substituting the given values:
\[ Z = \sqrt{(30)^2 + (40)^2} \]
\[ Z = \sqrt{900 + 1600} \]
\[ Z = \sqrt{2500} \]
\[ Z = 50 \, \Omega \]

### Step 3: Calculate the maximum current (I_max)
Using Ohm's law, the maximum current can be calculated using the formula:
\[ I_{max} = \frac{V_{max}}{Z} \]

Substituting the values:
\[ I_{max} = \frac{420 \, V}{50 \, \Omega} \]
\[ I_{max} = 8.4 \, A \]

### Final Answer
The maximum current is \( 8.4 \, A \).

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