A cell of emf 6 volts and internal resistance 2Ω is connected parallel with another cell of emf 9 volts and internal resistance 1Ω, when this arrangement is connected across a 12Ω resistance then the current through the resistance will be:

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:

• If the number of cells is 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:

• 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,

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

CALCULATION:

Given E₁ = 6 volts, r₁ = 2Ω, E₂ = 9 volts, r₂ = 1Ω, and R = 12Ω

If the parallel arrangement of n cells of emf E₁, E₂,..., E_n and internal resistance r1, r2,..., rn is connected across a resistance R, then the current in the resistance R is given as,

\(⇒ I=\frac{E_{eq}}{R+r_{eq}}\)     -----(1)

• The equivalent emf for the given arrangement is given as,

\(⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}\)

\(⇒ E_{eq}=\frac{6/2+9/1}{1/2+1/1}\)

⇒ E_eq = 8 volts

• The equivalent internal resistance for the given arrangement is given as,

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

\(⇒ \frac{1}{r_{eq}}=\frac{1}{2}+\frac{1}{1}\)

\(⇒ r_{eq}=\frac{2}{3}\Omega\)

By equation 1 the current through the resistance is given as,

\(⇒ I=\frac{E_{eq}}{R+r_{eq}}\)

\(⇒ I=\frac{8}{12+\frac{2}{3}}\)

\(⇒ I=\frac{24}{38}\,A\)

\(⇒ I=\frac{12}{19}\,A\)

• Hence, option 2 is correct.