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 a battery connected to three resistors in parallel, we can follow these steps:

### Step 1: Identify the Resistors and Their Values
We have three resistors connected in parallel:
- \( R_1 = 2 \, \Omega \)
- \( R_2 = 4 \, \Omega \)
- \( R_3 = 5 \, \Omega \)

### Step 2: Calculate the Effective Resistance
For resistors in parallel, the formula for the effective resistance \( R_{\text{effective}} \) is given by:
\[ \frac{1}{R_{\text{effective}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Substituting the values:
\[ \frac{1}{R_{\text{effective}}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} \]

### Step 3: Find a Common Denominator
The common denominator for \( 2, 4, \) and \( 5 \) is \( 20 \):
\[ \frac{1}{2} = \frac{10}{20}, \quad \frac{1}{4} = \frac{5}{20}, \quad \frac{1}{5} = \frac{4}{20} \]
Thus,
\[ \frac{1}{R_{\text{effective}}} = \frac{10}{20} + \frac{5}{20} + \frac{4}{20} = \frac{19}{20} \]

### Step 4: Calculate \( R_{\text{effective}} \)
Now, take the reciprocal to find \( R_{\text{effective}} \):
\[ R_{\text{effective}} = \frac{20}{19} \, \Omega \]

### Step 5: Use Ohm's Law to Find Total Current
The total current \( I \) drawn from the battery can be calculated using Ohm's Law:
\[ I = \frac{V}{R_{\text{effective}}} \]
Given that the EMF of the battery \( V = 20 \, V \):
\[ I = \frac{20}{\frac{20}{19}} = 20 \times \frac{19}{20} = 19 \, A \]

### Conclusion
The total current drawn from the battery is \( 19 \, A \).

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