A student is asked to connected four cells of e.m.f of 1 V and internal resistance 0.5 ohm in series with a external resistance of 1 ohm. But one cell is wrongly connected by him with its terminal reversed, the current in the circuit is

Options

A. `(1)/(3)A`
B. `(2)/(3)A`
C. `(3)/(4)A`
D. `(4)/(3)A`

Answer

To solve the problem, we need to analyze the circuit step by step.

### Step 1: Determine the Net EMF of the Circuit
In the circuit, we have four cells, each with an EMF of 1 V. However, one cell is connected in reverse. Therefore, the net EMF (E_net) can be calculated as follows:

- Three cells contribute positively: \(1 + 1 + 1 = 3 \, \text{V}\)
- One cell contributes negatively due to the reverse connection: \(-1 \, \text{V}\)

Thus, the total EMF is:
\[ E_{\text{net}} = 3 \, \text{V} - 1 \, \text{V} = 2 \, \text{V} \]

### Step 2: Calculate the Total Internal Resistance
Each cell has an internal resistance of \(0.5 \, \Omega\). Since there are four cells in series, the total internal resistance (R_internal) is:
\[ R_{\text{internal}} = 4 \times 0.5 \, \Omega = 2 \, \Omega \]

### Step 3: Determine the Total Resistance in the Circuit
The total resistance in the circuit (R_total) includes the external resistance (1 Ω) and the total internal resistance (2 Ω):
\[ R_{\text{total}} = R_{\text{external}} + R_{\text{internal}} = 1 \, \Omega + 2 \, \Omega = 3 \, \Omega \]

### Step 4: Calculate the Current in the Circuit
Using Ohm's law, the current (I) in the circuit can be calculated using the formula:
\[ I = \frac{E_{\text{net}}}{R_{\text{total}}} \]
Substituting the values we found:
\[ I = \frac{2 \, \text{V}}{3 \, \Omega} = \frac{2}{3} \, \text{A} \]

### Final Answer
The current in the circuit is:
\[ I = \frac{2}{3} \, \text{A} \]

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