Cells in Parallel: Equivalent EMF and Internal Resistance

Physics · Current Electricity · NEET

When cells are joined in parallel (all positive ends together, all negative ends together), the equivalent EMF is a weighted average: E_eq = (E1·r2 + E2·r1)/(r1 + r2), and the internal resistances combine like resistors in parallel: 1/r_eq = 1/r1 + 1/r2. Memory hook: "resistances add like parallel resistors; EMF is pulled toward the cell with the smaller internal resistance." This is a direct NCERT formula that NEET uses in short circuit and maximum-current numericals.
Two cells in parallel (positives joined, negatives joined)B1B2E1, r1E2, r2II1I2E_eq = (E1 r2 + E2 r1)/(r1 + r2), 1/r_eq = 1/r1 + 1/r2
Two cells connected in parallel: junction rule gives I = I1 + I2. Eliminating the branch currents yields the equivalent EMF (a weighted average) and an equivalent internal resistance that combines like parallel resistors.

Your doubts, answered

Is the equivalent EMF of parallel cells just E1 + E2?

No. Adding EMFs (E1 + E2) is the rule for cells in SERIES, not parallel. In parallel the EMF does not add up. It becomes a weighted average E_eq = (E1·r2 + E2·r1)/(r1 + r2). If both cells have the same EMF E, then E_eq = E (unchanged). Parallel gives you more current, not more voltage.

Why is r_eq smaller than each single internal resistance?

Because 1/r_eq = 1/r1 + 1/r2, so r_eq = (r1·r2)/(r1 + r2), which is always less than either r1 or r2. This is the same math as two resistors in parallel. Physically, the current now has two paths (two cells) to flow through, so the total opposition drops. That is the main benefit of parallel cells: lower internal resistance means the battery can supply more current.

What does E_eq become if both cells are identical (same E, same r)?

Put E1 = E2 = E and r1 = r2 = r. Then E_eq = (E·r + E·r)/(r + r) = 2Er/2r = E. And r_eq = r/2. So n identical cells in parallel give the SAME EMF E but internal resistance r/n. NEET loves this clean case: EMF stays E, internal resistance drops to r/n.

How do I remember which cell 'wins' in the EMF average?

Look at the formula E_eq = (E1·r2 + E2·r1)/(r1 + r2). Each EMF is weighted by the OTHER cell's resistance. So the cell with the smaller internal resistance dominates the result — E_eq is pulled toward it. A strong cell (small r) has more say in the combined EMF.

What if one cell's negative terminal joins the other's positive terminal?

Then that cell is reversed and its EMF enters with a minus sign (replace E2 with -E2 in the formula). NCERT states this directly. Always check the terminal connections before plugging into the formula, or you will get the sign wrong.

⚠️ The NEET trap
Adding the EMFs: E_eq = E1 + E2 for cells in parallel.
For parallel cells the EMF does NOT add. Use E_eq = (E1·r2 + E2·r1)/(r1 + r2), and 1/r_eq = 1/r1 + 1/r2. For n identical cells: E_eq = E and r_eq = r/n.
🧠 Series adds EMF (more voltage); parallel adds current paths (lower internal resistance). Never write E1 + E2 for a parallel combination.

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Frequently asked

What is the formula for equivalent EMF of two cells in parallel?

E_eq = (E1·r2 + E2·r1)/(r1 + r2), where E1, E2 are the EMFs and r1, r2 are the internal resistances of the two cells joined in parallel.

What is the equivalent internal resistance of cells in parallel?

They combine like resistors in parallel: 1/r_eq = 1/r1 + 1/r2, so r_eq = (r1·r2)/(r1 + r2). It is always smaller than either individual internal resistance.

For n identical cells in parallel, what are E_eq and r_eq?

E_eq = E (the same as one cell) and r_eq = r/n. The voltage is unchanged but the internal resistance drops by a factor of n, so the battery can deliver more current.

Why do we connect cells in parallel instead of series?

Parallel lowers the internal resistance and lets the battery supply a larger current without dropping much voltage. Series is used when you need a higher total EMF (voltage).

Does the general derivation come from Kirchhoff's junction rule?

Yes. NCERT starts from I = I1 + I2 (junction rule) and V = E1 - I1·r1 = E2 - I2·r2, then eliminates I1 and I2 to get E_eq and r_eq. It is a direct application of the junction rule.