A battery of four cells in series each having an efm of 1.5 V and internal resistance `1Omega` are connected in series with an ammeter, a coil of resistance `2Omega` and a filament lamp. If the ammeter reads 0.5 A, the resistance of the filament lamp is

Options

A. `4Omega`
B. `6Omega`
C. `2Omega`
D. `12Omega`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the total EMF and internal resistance of the battery
Given that there are 4 cells in series, each with an EMF of 1.5 V:
\[ E_{\text{total}} = 4 \times 1.5 \, \text{V} = 6 \, \text{V} \]
The internal resistance of each cell is 1 Ω, so for 4 cells in series:
\[ R_{\text{internal}} = 4 \times 1 \, \Omega = 4 \, \Omega \]

### Step 2: Write the total resistance in the circuit
The total resistance in the circuit includes the internal resistance of the battery, the resistance of the coil, and the resistance of the filament lamp (let's denote it as \( R \)):
\[ R_{\text{total}} = R_{\text{internal}} + R_{\text{coil}} + R_{\text{lamp}} = 4 \, \Omega + 2 \, \Omega + R \]
Thus,
\[ R_{\text{total}} = 6 \, \Omega + R \]

### Step 3: Apply Ohm's Law to find the relationship between voltage, current, and resistance
Using Ohm's Law, we know:
\[ V = I \times R_{\text{total}} \]
Substituting the known values:
\[ 6 \, \text{V} = 0.5 \, \text{A} \times (6 \, \Omega + R) \]

### Step 4: Solve for \( R \)
Rearranging the equation gives:
\[ 6 = 0.5 \times (6 + R) \]
Expanding this:
\[ 6 = 3 + 0.5R \]
Subtracting 3 from both sides:
\[ 3 = 0.5R \]
Multiplying both sides by 2:
\[ R = 6 \, \Omega \]

### Conclusion
The resistance of the filament lamp is \( R = 6 \, \Omega \).