An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω. The current through the resistance R is very close to 1 mA and the voltage across point A and B, V_AB = 1 V. Now the circuit is changed to : where value of V₂ = 5 V. The internal resistances of both the batteries are 0·1 Ω. The current through the resistance R is about :

Correct — C, about 3·0 mA. The printed figure shows the second circuit with the two batteries side by side, in parallel with one another, and that pair connected across the 1000 Ω resistor between terminals A and B.

Two cells of different EMF in parallel do not simply hand their larger voltage to the load. They settle at a common terminal voltage somewhere between the two EMFs, weighted by the internal resistances. With equal internal resistances of 0·1 Ω, that weighting is a plain average:

E_eq = (1/0·1 + 5/0·1) / (1/0·1 + 1/0·1) = 60 / 20 = 3 V, with r_eq = 0·1 ∥ 0·1 = 0·05 Ω.

The load is enormous compared with that internal resistance, so it barely matters:
I = 3 / (1000 + 0·05) ≈ 3·0 mA.

A node equation gives the same thing without the theorem: (1 − V)/0·1 + (5 − V)/0·1 = V/1000 solves to V = 2·99985 V across R, so I ≈ 3·0 mA.

Why the others are wrong

• (a)
  1·0 mA — This is the current in the FIRST circuit, before V₂ was added. Adding the second, higher-EMF cell raises the terminal voltage from 1 V to 3 V, so the current must rise too.
• (b)
  1·2 mA — This would need a terminal voltage of about 1·2 V, which would follow only if the 5 V cell barely influenced the node. With equal internal resistances it pulls exactly as hard as the 1 V cell, lifting the common voltage to the midpoint, 3 V.
• (d)
  5·0 mA — This assumes the larger battery simply wins and puts its full 5 V across R. It does not: the 1 V cell is still connected across the same two nodes and holds the terminal voltage down. The pair settles at 3 V, not 5 V.

Cells connected in parallel share a common pair of terminals, so they must share one terminal voltage. When their EMFs differ, that shared voltage is the weighted mean E_eq = ΣE/r ÷ Σ1/r, and the combination's internal resistance is the parallel combination of the individual internal resistances. This is Millman's theorem, and it is just Kirchhoff's current law applied at the shared node.

Key facts

• Parallel cells must share one terminal voltage.
• E_eq = (E1/r1 + E2/r2)/(1/r1 + 1/r2); with equal r this is the average of the EMFs.
• Here E_eq = (1 + 5)/2 = 3 V and r_eq = 0·05 ohm.
• I = E_eq/(R + r_eq) = 3/1000·05 ≈ 3·0 mA.
• The higher-EMF cell does not simply impose its voltage on the load.

Common traps

• Assuming the larger EMF simply appears across the load.
• Ignoring the lower-EMF cell because it seems 'weaker'.
• Treating parallel cells as if they were in series and adding the EMFs to 6 V.
• Forgetting that equal internal resistances make the result a simple average.

• practice — not a real PYQ

  Two cells of EMF 2 V and 6 V, each of internal resistance 1 Ω, are connected in parallel across a large resistance. The terminal voltage of the combination is about

  • (a)2 V
  • (b)4 V
  • (c)6 V
  • (d)8 V
  Answer(b) 4 V — with equal internal resistances the combination settles at the average of the two EMFs.
• practice — not a real PYQ

  Two identical cells each of EMF E are connected in series. The net EMF of the combination is

  • (a)E
  • (b)2E
  • (c)E/2
  • (d)zero
  Answer(b) 2E — EMFs add in series, unlike the parallel case where they average.