D.C motor run of 120 V . If armature current at `e_(b)` = 115 V is 10 A . Then find out current just after motor is started :

Options

A. 240 A
B. 230 A
C. 120 A
D. 400 A

Answer

To solve the problem, we need to find the current just after the DC motor is started, given the applied voltage, back EMF, and armature current at a certain condition. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Applied Voltage (V_applied) = 120 V
- Back EMF (E_b) when armature current (I_a) = 10 A is 115 V.

2. **Use Kirchhoff's Voltage Law (KVL):**
According to KVL for the DC motor circuit:
\[ V_{applied} = E_b + I_a \cdot R_a \]
Rearranging gives us:
\[ E_b = V_{applied} - I_a \cdot R_a \]

3. **Substitute Known Values:**
We know:
- \( V_{applied} = 120 \, V \)
- \( E_b = 115 \, V \)
- \( I_a = 10 \, A \)
Substitute these values into the equation:
\[ 115 = 120 - 10 \cdot R_a \]

4. **Solve for Armature Resistance (R_a):**
Rearranging the equation:
\[ 10 \cdot R_a = 120 - 115 \]
\[ 10 \cdot R_a = 5 \]
\[ R_a = \frac{5}{10} = 0.5 \, \Omega \]

5. **Determine Current Just After Starting:**
When the motor is just started, the back EMF (E_b) is zero because the motor is not yet in motion. Thus, we have:
\[ 0 = V_{applied} - I_a' \cdot R_a \]
Rearranging gives:
\[ I_a' \cdot R_a = V_{applied} \]
Substitute \( R_a = 0.5 \, \Omega \) and \( V_{applied} = 120 \, V \):
\[ I_a' \cdot 0.5 = 120 \]
\[ I_a' = \frac{120}{0.5} = 240 \, A \]

### Final Answer:
The current just after the motor is started is **240 A**.