The armature of a `DC` motor has `20Omega` resistance. It draws a current of `1.5 A` when run by `200 V DC` supply The value of back emf induced in it will be

Options

A. `150V`
B. `170V`
C. `180V`
D. `190V`

Explanation

To find the back EMF (Electromotive Force) induced in the armature of a DC motor, we can follow these steps:

### Step 1: Identify the given values
- Resistance of the armature (R) = 20 Ω
- Current (I) = 1.5 A
- Supply Voltage (V) = 200 V

### Step 2: Use Ohm's Law to calculate the voltage drop across the armature
According to Ohm's Law, the voltage drop (V_R) across the armature can be calculated using the formula:
\[ V_R = I \times R \]

Substituting the values:
\[ V_R = 1.5 \, \text{A} \times 20 \, \Omega = 30 \, \text{V} \]

### Step 3: Calculate the back EMF (E)
The relationship between the supply voltage (V), the back EMF (E), and the voltage drop across the armature (V_R) is given by:
\[ V = E + V_R \]

Rearranging this equation to solve for back EMF (E):
\[ E = V - V_R \]

Substituting the known values:
\[ E = 200 \, \text{V} - 30 \, \text{V} = 170 \, \text{V} \]

### Conclusion
The value of the back EMF induced in the DC motor is **170 V**.

---