Question

Ten cells, each having internal resistance \( 1\Omega \) and emf \( 1.5 \, V \) are connected in series. But unknowingly 3 cells are connected wrongly in series. The effective internal resistance and emf of the series combination are respectively:

• A. \( 0.1\Omega \) and \( 1.5 \, V \)
• B. \( 6\Omega \) and \( 10 \, V \)
• C. \( 10\Omega \) and \( 6 \, V \)
• D. \( 10\Omega \) and \( 15 \, V \)

Hint:

In series, resistances always add, but reversed cells reduce net emf by subtraction.

Answer 1

The Correct Option is C

Step 1: Total number of cells.
Total cells = 10, each with emf \( 1.5 \, V \).

Step 2: Identify wrongly connected cells.
3 cells are reversed, so their emf opposes the others.

Step 3: Calculate net emf.
Effective emf:
\[ E = (10 - 2 \times 3) \times 1.5 \]
\[ E = (10 - 6) \times 1.5 \]
\[ E = 4 \times 1.5 = 6 \, V \]

Step 4: Calculate internal resistance.
All internal resistances add in series:
\[ R = 10 \times 1 = 10\Omega \]

Step 5: Note important concept.
Reversing cells affects emf but not resistance.

Step 6: Final values.
\[ R = 10\Omega, \quad E = 6V \]

Step 7: Final Answer.
\[ \boxed{10\Omega \text{ and } 6V} \]