When n identical cells each of emf E are connected in series, the equivalent emf of the combination will be:

  1. nE
  2. \(\frac{E}{n}\)
  3. \(\frac{E}{n^2}\)
  4. n2E

Answer (Detailed Solution Below)

Option 1 : nE
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CUET General Awareness (Ancient Indian History - I)
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Detailed Solution

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CONCEPT:

Cell:

  • The cell converts chemical energy into electrical energy.
  • Cells are of two types:
    1. Primary cell: This type of cell cannot be recharged.
    2. Secondary cell: This type of cell can be recharged.
  • For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

Cells in series:

F1 P.B 8.5.20 Pallavi D7

  • If the number of cells are connected end to end that the positive terminal of one cell is connected to the negative terminal of the succeeding cell then it is called a series arrangement of cells.
  • The equivalent emf of cells in series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En

  • The equivalent internal resistance of cells in a series arrangement is given as,

⇒ req = r1 + r2 +...+ rn

Cells in parallel:

F1 P.B 8.5.20 Pallavi D8

  • If the number of cells is connected such that the positive terminals are connected together at one point and the negative terminals of these cells are connected together at another point then it is called a parallel arrangement of cells.
  • The equivalent emf of cells in a parallel arrangement is given as,

\(⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2+E_3/r_3+...+E_n/r_n}{1/r_1+1/r_2+1/r_3+...+1/r_n}\)

  • The equivalent internal resistance of cells in a parallel arrangement is given as,

\(⇒ \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...+\frac{1}{r_n}\)

EXPLANATION:

Give E1 = E2 = E3 = ... = En = E

We know that the equivalent emf of cells in a series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En

⇒ Eeq = E + E + ... + E

⇒ Eeq = nE

  • Hence, option 1 is correct.
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