Three resistances `2Omega, 4Omega, 5Omega` are combined in series and this combination is connected to a battery of 12 V emf and negligible internal resistance. The potential drop across these resistances are

Options

A. `(5,45,4.36,2.18)V`
B. `(2,18,5.45,4.36)V`
C. `(4.36,2.18,5.45)V`
D. `(2.18,4.36,5.45)V`

Explanation

To solve the problem of finding the potential drop across three resistances connected in series, we can follow these steps:

### Step 1: Calculate the Effective Resistance
When resistors are connected in series, the effective resistance \( R_{\text{eff}} \) is the sum of the individual resistances.

Given resistances:
- \( R_1 = 2 \, \Omega \)
- \( R_2 = 4 \, \Omega \)
- \( R_3 = 5 \, \Omega \)

The effective resistance is calculated as:
\[ R_{\text{eff}} = R_1 + R_2 + R_3 = 2 + 4 + 5 = 11 \, \Omega \]

### Step 2: Calculate the Total Current
Using Ohm's Law, the total current \( I \) flowing through the circuit can be calculated using the formula:
\[ I = \frac{V}{R_{\text{eff}}} \]
Where \( V = 12 \, V \) (the voltage of the battery).

Substituting the values:
\[ I = \frac{12 \, V}{11 \, \Omega} = \frac{12}{11} \, \text{A} \approx 1.09 \, \text{A} \]

### Step 3: Calculate the Potential Drop Across Each Resistor
Using Ohm's Law \( V = I \times R \), we can calculate the potential drop across each resistor.

#### For \( R_1 = 2 \, \Omega \):
\[ V_1 = I \times R_1 = \left(\frac{12}{11}\right) \times 2 = \frac{24}{11} \, V \approx 2.18 \, V \]

#### For \( R_2 = 4 \, \Omega \):
\[ V_2 = I \times R_2 = \left(\frac{12}{11}\right) \times 4 = \frac{48}{11} \, V \approx 4.36 \, V \]

#### For \( R_3 = 5 \, \Omega \):
\[ V_3 = I \times R_3 = \left(\frac{12}{11}\right) \times 5 = \frac{60}{11} \, V \approx 5.45 \, V \]

### Step 4: Summarize the Results
The potential drops across the resistors are:
- \( V_1 \approx 2.18 \, V \)
- \( V_2 \approx 4.36 \, V \)
- \( V_3 \approx 5.45 \, V \)

### Final Answer
The potential drops across the resistances are:
- Across \( 2 \, \Omega \): \( 2.18 \, V \)
- Across \( 4 \, \Omega \): \( 4.36 \, V \)
- Across \( 5 \, \Omega \): \( 5.45 \, V \)

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