Efficiency and Thermal Dissipation in Non-Ideal Batteries: Calculating the maximum power transfer, circuit efficiency, and thermal dissipation when multiple non-identical cells (with different EMFs and internal resistances) are connected in parallel.

1. Basic Idea

A real battery is not an ideal source of emf. It has an internal resistance. If its emf is E and internal resistance is r, then when current I flows, some electrical energy is converted into heat inside the battery.

V = E - Ir

where:

• E = emf of the battery
• r = internal resistance
• I = current
• V = terminal voltage

When several non-identical batteries are connected in parallel, they can drive current through one another because their emfs and internal resistances are different.

2. Two Non-Identical Cells in Parallel

Consider two cells connected in parallel:

• Cell 1: emf E₁, internal resistance r₁
• Cell 2: emf E₂, internal resistance r₂

Let their common terminal voltage be V.

Current supplied by cell 1:

I₁ = (E₁ - V)/r₁

Current supplied by cell 2:

I₂ = (E₂ - V)/r₂

If an external load R is connected, load current is:

I = V/R

Using Kirchhoff's current law:

I₁ + I₂ = I

Therefore:

(E₁-V)/r₁ + (E₂-V)/r₂ = V/R

3. Equivalent EMF and Internal Resistance

For multiple cells connected in parallel, the equivalent internal resistance is:

1/r_eq = 1/r₁ + 1/r₂ + ... + 1/r_n

The equivalent emf is:

E_eq = [ E₁/r₁ + E₂/r₂ + ... + E_n/r_n ] / [ 1/r₁ + 1/r₂ + ... + 1/r_n ]

Important: The equivalent emf is a conductance-weighted average of the individual emfs. The cell having smaller internal resistance has greater influence.

4. Formula for Two Parallel Cells

For two cells:

E_eq = (E₁r₂ + E₂r₁)/(r₁+r₂)

and:

r_eq = r₁r₂/(r₁+r₂)

If an external resistance R is connected:

I = E_eq/(R+r_eq)

Terminal voltage:

V=IR

5. Maximum Power Transfer

For an equivalent battery having emf E_eq and internal resistance r_eq, connected to an external resistance R:

P = I²R

Since:

I=E_eq/(R+r_eq)

Therefore:

P = E_eq²R/(R+r_eq)²

Maximum power is obtained when:

R = r_eq

Therefore:

P_max = E_eq²/(4r_eq)

6. Circuit Efficiency

Useful power delivered to the external load:

P_load=I²R

Total power generated by the ideal emf source:

P_total=EI

Therefore efficiency is:

η = P_load/P_total

Hence:

η = R/(R+r)

or:

η = V/E

At maximum power:

R=r

Therefore:

η=50%

Maximum power transfer does NOT mean maximum efficiency. At maximum power transfer, efficiency is only 50%.

7. Thermal Dissipation

The power dissipated as heat in internal resistance is:

P_heat=I²r

For multiple parallel cells:

P_internal = I₁²r₁ + I₂²r₂ + ... + I_n²r_n

This heat represents energy lost inside the batteries.

When non-identical cells are connected in parallel without a load, the higher-emf cell can force current through the lower-emf cell. This causes internal heating and energy loss.

8. Circulating Current Between Two Cells

Suppose two cells are connected in parallel without an external load. If E₁ > E₂, current circulates from the higher-emf cell toward the lower-emf cell.

The circulating current is:

I_c = (E₁-E₂)/(r₁+r₂)

The heat generated in the two internal resistances is:

P_heat = I_c²(r₁+r₂)

Therefore:

P_heat = (E₁-E₂)²/(r₁+r₂)

This is an important competitive-exam result.

9. Solved Example 1

Two cells of emf 12 V and 10 V have internal resistances 1 Ω and 2 Ω. They are connected in parallel. Find their equivalent emf and internal resistance.

Solution

E_eq = (E₁r₂+E₂r₁)/(r₁+r₂)

E_eq = (12×2+10×1)/(1+2)

E_eq=34/3

E_eq = 11.33 V

Equivalent resistance:

r_eq=1×2/(1+2)

r_eq=2/3 Ω

10. Solved Example 2 — Maximum Power

An equivalent battery has E = 12 V and internal resistance r = 3 Ω. Find maximum power delivered to the external load.

P_max=E²/(4r)

P_max=144/(12)

P_max=12 W

For maximum power:

R=r=3 Ω

11. Solved Example 3 — Circulating Current

Two cells of emf 10 V and 8 V have internal resistances 1 Ω and 3 Ω. They are connected in parallel without an external load. Find the circulating current.

I_c=(E₁-E₂)/(r₁+r₂)

I_c=(10-8)/(1+3)

I_c=0.5 A

Heat dissipated:

P=I²(r₁+r₂)

P=(0.5)²×4

P=1 W

12. Solved Example 4 — Efficiency

A battery has emf 20 V and internal resistance 2 Ω. It supplies a current of 4 A to an external load. Find the efficiency.

Using:

η=1-r/(R+r)

Since:

R=E/I-r

R=20/4-2=3 Ω

Therefore:

η=R/(R+r)

η=3/5

η=60%

13. Important Formula Sheet

Quantity	Formula
Equivalent resistance	1/req=Σ(1/ri)
Equivalent emf	Eeq=Σ(Ei/ri) / Σ(1/ri)
Load current	I=Eeq/(R+req)
Load power	P=I²R
Maximum power condition	R=req
Maximum power	Pmax=Eeq²/(4req)
Efficiency	η=R/(R+req)
Internal heat	Pheat=I²r
Circulating current	Ic=(E₁-E₂)/(r₁+r₂)

Section A — 1 Mark
10 MCQs

Q1. The terminal voltage of a cell delivering current I is: 1 Mark

A) E+Ir

B) E-Ir

C) Ir-E

D) E/r

Answer: B

Q2. Maximum power is delivered to the load when: 1 Mark

A) R=0

B) R=∞

C) R=r

D) R=2r

Answer: C

Q3. Efficiency at maximum power transfer is: 1 Mark

A) 25%

B) 50%

C) 75%

D) 100%

Answer: B

Q4. Heat produced in an internal resistance r is: 1 Mark

A) Ir

B) I²r

C) I/r

D) Er

Answer: B

Q5. For two cells in parallel, equivalent internal resistance is: 1 Mark

A) r₁+r₂

B) r₁-r₂

C) r₁r₂/(r₁+r₂)

D) r₁/r₂

Answer: C

Q6. When two unequal-emf cells are connected in parallel, current may: 1 Mark

A) Become zero always

B) Circulate between cells

C) Become infinite

D) Reverse emf

Answer: B

Q7. Equivalent emf of parallel non-identical cells is a: 1 Mark

A) Simple arithmetic sum

B) Conductance-weighted average

C) Difference

D) Product

Answer: B

Q8. Load power is given by: 1 Mark

A) I²R

B) IR

C) I/R

D) R/I

Answer: A

Q9. Internal power loss of a battery is: 1 Mark

A) I²r

B) I²R

C) ER

D) E/R

Answer: A

Q10. A battery with smaller internal resistance has: 1 Mark

A) Less influence in parallel combination

B) Greater influence in equivalent emf

C) Zero emf

D) Infinite resistance

Answer: B

Section B — 2 Marks
10 Questions

Q11. State the condition for maximum power transfer. 2 Marks

R = r_eq, where R is external load resistance.

Q12. Write the efficiency of a battery having internal resistance r and load resistance R. 2 Marks

η=R/(R+r)

Q13. Two cells of 6 V and 4 V have internal resistances 1 Ω and 1 Ω. Find their equivalent emf when connected in parallel. 2 Marks

E_eq=(6×1+4×1)/(1+1)

E_eq=5 V

Q14. Find equivalent internal resistance of two cells having internal resistances 2 Ω and 6 Ω in parallel. 2 Marks

r_eq=2×6/(2+6)

r_eq=1.5 Ω

Q15. A battery has E=10 V and r=2 Ω. Find maximum power. 2 Marks

P_max=10²/(4×2)

P_max=12.5 W

Q16. Find efficiency when R=8 Ω and r=2 Ω. 2 Marks

η=8/(8+2)=0.8

η=80%

Q17. Why does a non-identical parallel combination produce internal heating? 2 Marks

Because unequal emfs produce a circulating current between the cells. This current causes Joule heating I²r in their internal resistances.

Q18. Two cells of 12 V and 10 V have internal resistances 2 Ω and 3 Ω. Find the circulating current when connected without load. 2 Marks

I_c=(12-10)/(2+3)

I_c=0.4 A

Q19. What is the internal power loss when I=3 A and r=2 Ω? 2 Marks

P=I²r=9×2

P=18 W

Q20. Explain why maximum power and maximum efficiency occur at different conditions. 2 Marks

Maximum power occurs when R=r, giving 50% efficiency. Efficiency increases as R becomes much larger than r, approaching 100% but with very small current and hence low power.

Section C — 3 Marks
10 CBSE + Competitive Questions

Q21. Derive the equivalent emf of two non-identical cells connected in parallel. 3 Marks

Let common terminal voltage be V. Current from cell 1:

I₁=(E₁-V)/r₁

Current from cell 2:

I₂=(E₂-V)/r₂

The equivalent source satisfies:

V=E_eq-Ir_eq

Using conductance weighting:

E_eq = (E₁/r₁+E₂/r₂)/(1/r₁+1/r₂)

Hence:

E_eq=(E₁r₂+E₂r₁)/(r₁+r₂)

Q22. Two cells of 12 V, 1 Ω and 6 V, 2 Ω are connected in parallel. Find equivalent emf and internal resistance. 3 Marks

E_eq=(12×2+6×1)/(1+2)

E_eq=30/3=10 V

r_eq=1×2/(1+2)=2/3 Ω

E_eq=10 V, r_eq=2/3 Ω

Q23. The equivalent battery of Q22 is connected to a 10 Ω load. Find current and terminal voltage. 3 Marks

I=10/(10+2/3)

I≈0.9375 A

Terminal voltage:

V=IR

V≈0.9375×10

V≈9.375 V

Q24. Find the maximum power delivered by the equivalent battery of Q22. 3 Marks

P_max=E²/(4r)

P_max=100/[4×(2/3)]

P_max=37.5 W

Q25. Calculate the efficiency of the battery in Q23. 3 Marks

η=R/(R+r)

η=10/(10+2/3)

η≈93.75%

Q26. Two cells of emf 15 V and 10 V with internal resistances 2 Ω and 3 Ω are connected in parallel without external load. Find circulating current and total heat loss. 3 Marks

I_c=(15-10)/(2+3)

I_c=1 A

Heat loss:

P=I²(r₁+r₂)

P=1²×5

I_c=1 A, P_heat=5 W

Q27. Derive the expression for maximum power delivered by a non-ideal battery. 3 Marks

Current:

I=E/(R+r)

Load power:

P=I²R

Therefore:

P=E²R/(R+r)²

Differentiating with respect to R and putting dP/dR=0 gives:

R=r

Therefore:

P_max=E²/(4r)

Q28. Three identical cells each have emf E and internal resistance r. They are connected in parallel. Find equivalent emf and internal resistance. 3 Marks

For identical cells:

E_eq=E

Internal resistances are parallel:

1/r_eq=3/r

Therefore:

r_eq=r/3

Thus:

E_eq=E,   r_eq=r/3

Q29. Two parallel cells have emfs E₁ and E₂ and internal resistances r₁ and r₂. Show that the higher-emf cell supplies current while the lower-emf cell may absorb current. 3 Marks

Let common terminal voltage be V. For cell 1:

I₁=(E₁-V)/r₁

For cell 2:

I₂=(E₂-V)/r₂

If:

E₁>V>E₂

then:

I₁>0

but:

I₂<0

Therefore cell 1 supplies current while cell 2 absorbs current and gets charged internally.

Q30. A battery combination has equivalent emf 24 V and internal resistance 4 Ω. Find the load resistance for maximum power, maximum power and efficiency at maximum power. 3 Marks

For maximum power:

R=r=4 Ω

Maximum power:

P_max=24²/(4×4)

P_max=36 W

Efficiency:

η=R/(R+r)

η=4/(4+4)=1/2

Load resistance = 4 Ω
Maximum power = 36 W
Efficiency = 50%

Final Exam Revision

E_eq = Σ(E_i/r_i) / Σ(1/r_i)

1/r_eq=Σ1/r_i

I=E_eq/(R+r_eq)

P_load=I²R

P_heat=I²r_eq

R=r_eq → P_max

P_max=E_eq²/(4r_eq)

η=R/(R+r_eq)

η at maximum power = 50%

I_c=(E₁-E₂)/(r₁+r₂)

Competitive Exam Trick:

When non-identical cells are connected in parallel, do NOT simply average their emfs. The cell with smaller internal resistance gets greater weight.

If there is no external load and E₁ ≠ E₂, expect a circulating current and therefore thermal dissipation inside the cells.

Efficiency and Thermal Dissipation in Non-Ideal Batteries
CBSE + Competitive Physics | 30 Questions | 1, 2 & 3 Marks