Three resistors `2Omega, 4Omega and 5Omega` are combined in parallel. This combination is connected to a battery of emf 20V and negligible internal Resistance. The total current drawn from the battery is

Options

A. 10A
B. 15A
C. 19A
D. 23A

Explanation

To solve the problem of finding the total current drawn from the battery when three resistors (2Ω, 4Ω, and 5Ω) are connected in parallel to a 20V battery, we can follow these steps:

### Step 1: Calculate the Equivalent Resistance (R_eq) of the Resistors in Parallel
The formula for calculating the equivalent resistance (R_eq) for resistors in parallel is given by:

\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

Where:
- \( R_1 = 2Ω \)
- \( R_2 = 4Ω \)
- \( R_3 = 5Ω \)

Substituting the values:

\[ \frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} \]

### Step 2: Find a Common Denominator
To add the fractions, we need a common denominator. The least common multiple (LCM) of 2, 4, and 5 is 20.

### Step 3: Rewrite Each Fraction
Now convert each fraction to have a denominator of 20:

\[ \frac{1}{2} = \frac{10}{20}, \quad \frac{1}{4} = \frac{5}{20}, \quad \frac{1}{5} = \frac{4}{20} \]

### Step 4: Add the Fractions
Now, add the fractions:

\[ \frac{1}{R_{eq}} = \frac{10}{20} + \frac{5}{20} + \frac{4}{20} = \frac{19}{20} \]

### Step 5: Calculate R_eq
Taking the reciprocal to find \( R_{eq} \):

\[ R_{eq} = \frac{20}{19} \, \Omega \]

### Step 6: Use Ohm's Law to Find the Total Current (I)
Using Ohm's Law, the total current drawn from the battery can be calculated using the formula:

\[ I = \frac{V}{R_{eq}} \]

Where:
- \( V = 20V \)

Substituting the values:

\[ I = \frac{20V}{\frac{20}{19} \, \Omega} \]

### Step 7: Simplify the Expression
This simplifies to:

\[ I = 20V \times \frac{19}{20} = 19A \]

### Conclusion
The total current drawn from the battery is **19A**.

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