$ ext{Six lead acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 } \Omega ext{ are connected in series to provide a supply to a resistance of 8.5 } \Omega ext{. The value of current drawn from supply will be :}$

Q: $\text{Six lead acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 } \Omega \text{ are connected in series to provide a supply to a resistance of 8.5 } \Omega \text{. The value of current drawn from supply will be :}$

$\text{Six lead acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 } \Omega \text{ are connected in series to provide a supply to a resistance of 8.5 } \Omega \text{. The value of current drawn from supply will be :}$

Question

$	ext{Six lead acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 } \Omega 	ext{ are connected in series to provide a supply to a resistance of 8.5 } \Omega 	ext{. The value of current drawn from supply will be :}$

Accepted Answer

To determine the current drawn from the supply, we need to calculate the total voltage provided by the cells and the total resistance in the circuit.

1. **Total Voltage:**
   - Each lead acid cell has an emf of 2.0 V.
   - There are six cells connected in series, so the total emf (E) is:
     $$
     E = 6 \times 2.0 \, \text{V} = 12.0 \, \text{V}
     $$

2. **Total Internal Resistance:**
   - Each cell has an internal resistance of 0.015 Ω.
   - Since the cells are in series, the total internal resistance (r) is:
     $$
     r = 6 \times 0.015 \, \Omega = 0.09 \, \Omega
     $$

3. **Total Resistance in the Circuit:**
   - The external resistance (R) is given as 8.5 Ω.
   - Therefore, the total resistance (R_total) in the circuit is the sum of the external resistance and the total internal resistance:
     $$
     R_{\text{total}} = R + r = 8.5 \, \Omega + 0.09 \, \Omega = 8.59 \, \Omega
     $$

4. **Current Calculation:**
   - Using Ohm's Law, the current (I) drawn from the supply is given by:
     $$
     I = \frac{E}{R_{\text{total}}} = \frac{12.0 \, \text{V}}{8.59 \, \Omega} \approx 1.40 \, \text{A}
     $$

Therefore, the correct answer is Option 4: 1.40 A.

**Explanation for Incorrect Options:**

- **Option 1 (0.24 A):** This value is significantly lower than the calculated current. It does not account for the correct total voltage and resistance in the circuit.
  
- **Option 2 (0.72 A):** This value is approximately half of the correct current. It might result from an incorrect calculation or misunderstanding of the series connection of cells and resistances.
  
- **Option 3 (1.20 A):** This value is close to the correct answer but still incorrect. It could result from a minor calculation error or incorrect rounding, but it does not match the precise calculation based on the given values.

In conclusion, the detailed calculation confirms that Option 4 (1.40 A) is indeed the correct answer, as it accurately reflects the current drawn from the supply based on the given parameters.