While connecting 6 cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective resistance of load is 24 ohm and internal resistance of each cell is one ohm and emf 1.5 V the current delivered by the battery is

Options

A. 0.1 A
B. 0.2 A
C. 0.3 A
D. 0.4 A

Explanation

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

### Step 1: Determine the effective EMF of the battery
We have 6 cells connected in series, but one cell is connected with reverse polarity. Each cell has an EMF (E) of 1.5 V.

- **Effective EMF (E_eff)** from 5 cells connected normally and 1 cell connected in reverse:
\[ E_{\text{eff}} = 5E - E = 5(1.5) - 1.5 = 7.5 - 1.5 = 6 \text{ V} \]

### Step 2: Calculate the total internal resistance
Each cell has an internal resistance (r) of 1 ohm. Since there are 6 cells, the total internal resistance (R_internal) is:
\[ R_{\text{internal}} = 6r = 6 \times 1 = 6 \text{ ohm} \]

### Step 3: Determine the total resistance in the circuit
The total resistance in the circuit (R_total) is the sum of the internal resistance and the load resistance (R_load):
- Given load resistance \( R_{\text{load}} = 24 \text{ ohm} \)

Thus, the total resistance is:
\[ R_{\text{total}} = R_{\text{internal}} + R_{\text{load}} = 6 + 24 = 30 \text{ ohm} \]

### Step 4: Calculate the current delivered by the battery
Using Ohm's Law, the current (I) can be calculated using the formula:
\[ I = \frac{E_{\text{eff}}}{R_{\text{total}}} \]
Substituting the values we found:
\[ I = \frac{6 \text{ V}}{30 \text{ ohm}} = 0.2 \text{ A} \]

### Final Answer
The current delivered by the battery is **0.2 A**.

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