A student connects four cells, each of emf 1.5 V and internal resistance `0.25 Omega` in series but one cell has its terminals reversed. This battery sends current in a `2Omega` resistor . What will be the current ?

Options

A. 2A
B. 4A
C. 1A
D. 3A

Answer

To solve the problem, we need to find the current flowing through a circuit consisting of four cells connected in series, with one cell having its terminals reversed. Here’s a step-by-step solution:

### Step 1: Determine the total EMF of the battery
Each cell has an EMF of 1.5 V. Since there are four cells, the total EMF would normally be:

\[ \text{Total EMF} = 1.5 \, \text{V} + 1.5 \, \text{V} + 1.5 \, \text{V} + 1.5 \, \text{V} = 6 \, \text{V} \]

However, since one cell is reversed, we subtract the EMF of that cell:

\[ \text{Net EMF} = 6 \, \text{V} - 1.5 \, \text{V} = 4.5 \, \text{V} \]

### Step 2: Calculate the total internal resistance
Each cell has an internal resistance of \(0.25 \, \Omega\). Since all four cells are in series, their internal resistances add up:

\[ \text{Total Internal Resistance} = 0.25 \, \Omega + 0.25 \, \Omega + 0.25 \, \Omega + 0.25 \, \Omega = 1 \, \Omega \]

### Step 3: Add the external resistance
The external resistor is given as \(2 \, \Omega\). Therefore, the total resistance in the circuit is:

\[ \text{Total Resistance} = \text{Total Internal Resistance} + \text{External Resistance} = 1 \, \Omega + 2 \, \Omega = 3 \, \Omega \]

### Step 4: Calculate the current using Ohm's Law
Using Ohm's Law, the current \(I\) can be calculated as:

\[ I = \frac{\text{Net EMF}}{\text{Total Resistance}} = \frac{4.5 \, \text{V}}{3 \, \Omega} = 1.5 \, \text{A} \]

### Final Answer
The current flowing through the circuit is \(1.5 \, \text{A}\).

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