Q.Two cells are connected in opposition to each other, forming a single closed loop with no external resistor. Cell E1 has emf 6 V and internal resistance 2 Ω; cell E2 has emf 4 V and internal resistance 8 Ω. Their emfs act against each other around the loop, and A and B are the two junction points at which the two cells meet. Find the potential difference between the points A and B.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Internal Resistance
Internal Resistance – The Battery That Fights Itself
Imagine you have a bucket of water with a tap at the bottom. When you open the tap fully, water flows out freely. But if the pipe is narrow or clogged, the flow is weaker even though the bucket is full. A battery behaves the same way: it has a "full bucket" of electrical energy (its emf, or electromotive force), but inside the battery, there is always some "clogged pipe" that resists the flow of charge. That clog is called internal resistance.
The Intuition
Every real battery is not a perfect source of voltage. If you connect a small bulb to a fresh battery, it glows brightly. But if you connect a powerful motor that draws a large current, the same battery might struggle — the voltage at its terminals drops, and the motor runs slower. Why? Because inside the battery, the chemical reactions and the materials themselves have a natural opposition to the flow of charge. This opposition is internal resistance, denoted by r (or sometimes Rint).
Think of it this way: the battery has two "battles" to fight. First, it must push charge through the external circuit (the bulb, the motor, the wires). Second, it must push charge through its own internal structure. The harder it has to push (i.e., the larger the current), the more voltage it "loses" inside itself.
The Precise Statement
Internal resistance is the opposition to the flow of electric current offered by the materials and chemical processes inside a source of emf (like a battery, cell, or generator). It is measured in ohms (Ω).
For a cell or battery, the relationship between the emf (E), the terminal voltage (V), the current (I), and the internal resistance (r) is given by:
V=E−Ir
This is the single most important equation to remember.
V=E−Ir
Here:
- E is the electromotive force — the maximum voltage the battery can provide when no current flows (open circuit). Think of it as the "ideal" voltage.
- V is the terminal voltage — the actual voltage you measure across the battery's terminals when it is delivering current.
- I is the current flowing through the circuit (and therefore through the battery itself).
- r is the internal resistance.
The term Ir is the voltage drop inside the battery. It is the "price" the battery pays for delivering current.
What Happens in Different Situations?
| Condition | Current I | Terminal Voltage V | Why? |
|---|---|---|---|
| Open circuit (no load) | I=0 | V=E | No current means no internal drop. |
| Small load (e.g., a dim bulb) | Small I | V≈E | The Ir term is tiny. |
| Large load (e.g., a powerful motor) | Large I | V<E significantly | The Ir drop becomes noticeable. |
| Short circuit (wires directly across terminals) | Very large I | V≈0 | Almost all voltage is lost inside the battery; the battery heats up and may be damaged. |
A common mistake is to think that internal resistance is a "bad" thing that can be eliminated. It cannot — every real source has some internal resistance. Even a brand new AA battery has a small r (typically 0.1 to 0.5 Ω). Old or weak batteries have much larger r, which is why they fail to power high-current devices.
Why Does Internal Resistance Matter?
- Power loss inside the battery: The power dissipated as heat inside the battery is Ploss=I2r. This is why batteries get warm when delivering large currents.
- Maximum power transfer: There is a famous theorem (Maximum Power Transfer Theorem) that says a source delivers maximum power to an external load when the load resistance equals the internal resistance (Rload=r). But this is inefficient — half the power is wasted inside the source. …
Why this formula?
Internal Resistance: Why the Key Formulas Hold
Internal resistance is a fundamental concept in real-world circuits. No battery is perfect — every practical source has some internal resistance (r) that opposes the flow of current inside the source itself.
1. The Core Idea: A Real Battery = An Ideal Source + A Resistor
Think of a real battery as:
- An ideal EMF source (E) — provides constant voltage with zero internal resistance.
- A small resistor (r) — connected in series inside the battery.
Why series? Because the current that leaves the battery must first pass through its internal material (electrolyte, electrodes), which offers resistance.
2. The Terminal Voltage Formula
When the battery delivers current I to an external circuit:
- The ideal source produces E.
- The internal resistor r drops some voltage: Vdrop=Ir (Ohm's law).
- The voltage available at the terminals (V) is what's left:
V=E−Ir
Why this makes sense:
- If I=0 (open circuit), V=E — you measure the full EMF.
- If I increases, V decreases — the internal drop grows.
- If I is huge (short circuit), V→0 and all voltage is lost inside.
3. The Short-Circuit Current Formula
If you connect the terminals directly (external resistance R=0):
- The only resistance in the circuit is r.
- By Ohm's law: Ishort=rE
Why this holds:
- The entire EMF is now dropped across r alone.
- This is the maximum current the battery can deliver — limited by its internal resistance.
- In practice, this can damage the battery (heating, chemical damage).
4. Power Delivered to External Load
When the battery is connected to an external load R:
- Total circuit resistance: R+r
- Current: I=R+rE
- Power delivered to the external load (Pout):
Pout=I2R=(R+rE)2R
Why this matters:
- Power is not simply E2/R — because r limits current. …
The cells oppose, so the net emf driving the loop is 6−4=2 V across a total resistance 2+8=10 Ω, giving a loop current of 0.2 A. The terminal potential difference between A and B is then 5.6 V. …
Because the two cells are connected in opposition, only their difference in emf drives current. The net emf 6−4=2 V pushes a current of 0.2 A through the total internal resistance 2+8=10 Ω. Evaluating the terminal voltage of either cell gives the potential difference across A and B as 5.6 V.
Concept
With no external resistor, the two cells form one loop. Being in opposition, the effective emf is E1−E2 and the total resistance is the sum of the internal resistances.
Step 1 — loop current
I=r1+r2E1−E2=2+86−4=102=0.2 A.
The current is driven by the stronger cell E1; the weaker cell E2 is being charged (current is forced through it against its emf).
Step 2 — potential difference between A and B …
Method: Two Cells Opposing Each Other in a Single Closed Loop
Use this for any circuit where two cells are connected so their emfs act against each other around one loop, with no external resistor — a common setup for finding the potential difference between their two junction points.
Steps
Step 1: Identify the net driving emf
When two cells oppose each other, only the difference of their emfs drives current — the stronger cell sets the direction:
I=r1+r2E1−E2
(take E1>E2 so I comes out positive, driven by cell 1.)
Step 2: Recognise which cell is discharging and which is being charged
The stronger cell (E1) discharges — delivers current in the direction its own emf favours. The weaker cell (E2) has current forced backward through it against its own emf, so it is being charged.
Step 3: Compute the junction-to-junction voltage from EITHER cell, as a built-in check …
- GUJCET 2026Set x1 markMCQQ.The emf of a storage battery of a car is 6.0 V. If internal resistance of the battery is 0.2 Ω, then maximum power drawn from the battery is ____ W. (A) 2.4 (B) 180 (C) 15 (D) Zero
›Reveal solutionSolution
Maximum power drawn from the battery (short circuit) =ε2/r=180 W.
The total power delivered by the source is maximum when the current is maximum, i.e. under short circuit (Rext=0): …
- GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQQ.The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.8 Ohm, what is the maximum current that can be drawn from the battery?(a) 15 A(b) 1.5 A(c) 0.15 A(d) 30 A
›Reveal solutionSolution
The maximum current a battery can deliver occurs when the external resistance is zero (short circuit), so the entire emf drives current through only the internal resistance: I_max = emf/r.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.6Ω then the maximum current that can be drawn from the battery is ___.(a) 20 A(b) 25 A(c) 30 A(d) 72 A
›Reveal solutionSolution
The maximum current a cell/battery can deliver occurs on short circuit, where I_max = ε/r.
ε = 12 V, r = 0.6 Ω.
…
- GUJCET 2023Set 091 markMCQQ.Figure shows 2.0 V potentiometer used for the determination of internal resistance of 1.5 V cell. The balance point of cell in open circuit is 77.4 cm. When a resistor of 9.6 Ω is used in the external circuit of the cell, the balanced point shifts to 64.5 cm length of the potentiometer wire. The internal resistance of the cell is ______ Ω. (A) 1.92 (B) 1.5 (C) 1.62 (D) 0.96
›Reveal solutionSolution
Potentiometer internal-resistance formula: r = R(l₁ − l₂)/l₂.
Concept. With balance length l1 in open circuit (EMF) and l2 when an external resistor R is across the cell, r=R(l2l1−l2).
Solution. l1=77.4 cm, l2=64.5 cm, R=9.6Ω: …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.Two batteries of emf epsilon_1 & epsilon_2 (epsilon_2 > epsilon_1) and internal resistance r_1 & r_2 respectively are connected in parallel as shown.(a) The equivalent emf is given by epsilon_eq = epsilon_1 + epsilon_2(b) The equivalent emf epsilon_eq is smaller than epsilon_1(c) The equivalent emf epsilon_eq of the two cells is between epsilon_1 & epsilon_2 i.e. epsilon_1 < epsilon_eq < epsilon_2(d) epsilon_eq is independent of internal resistance r_1 & r_2
›Reveal solutionSolution
Combining two cells in parallel gives epsilon_eq = (epsilon_1 r_2 + epsilon_2 r_1)/(r_1 + r_2), a weighted mean, so epsilon_1 < epsilon_eq < epsilon_2.
For two cells in parallel, the equivalent emf and internal resistance are:
epsilon_eq = (epsilon_1 r_2 + epsilon_2 r_1)/(r_1 + r_2),
1/r_eq = 1/r_1 + 1/r_2.
The expression for epsilon_eq is a weighted average of epsilon_1 and epsilon_2 (weights r_2 and r_1). A weighted average of two numbers always lies between them:
epsilon_1 < epsilon_eq < epsilon_2 (since epsilon_2 > epsilon_1).
…
- GUJCET 2020Set 071 markMCQQ.The emf of a car battery is 12V of internal resistance of battery is 0.4Ω then maximum power drawn from battery is ______ W. (A) 4.8 (B) 360 (C) 30 (D) Zero
›Reveal solutionSolution
The maximum total power the source can supply is at short circuit, P=E2/r=144/0.4=360 W.
Concept. The total power delivered by the EMF is P=EI=R+rE2, which is largest when the external resistance R→0 (maximum current, i.e. short circuit).
Compute. …
- GUJCET 2020Set 071 markMCQQ.One electric cell (having emf of 2V & internal resistance of 0.1Ω) and other electric cell (having emf of 4V & internal resistance of 0.2Ω) are connected in parallel to each other. Then its equivalent emf will be ______ V. (A) 1.33 (B) 2.57 (C) 2.67 (D) 0.38
›Reveal solutionSolution
For cells in parallel, Eeq=1/r1+1/r2E1/r1+E2/r2=1540=2.67 V.
Concept. Combining two cells in parallel, the equivalent emf is the conductance-weighted average:
Eeq=r11+r21r1E1+r2E2. …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.Figure shows a part of a closed circuit. If the current flowing through it is 2A. What will be the potential difference between points B to A?(a) +2V(b) +1V(c) -1V(d) -2V
›Reveal solutionSolution
Adding the potential changes along the path from A to B - two resistor drops of I*R = 2 x 0.25 = 0.5 V each plus the 1 V cell - gives V_B - V_A = -2 V.
Current I = 2 A, each resistor R = 1/4 ohm.
Drop across each resistor in the direction of current: I*R = 2 x 0.25 = 0.5 V (potential falls).
Traversing A -> B, the potential falls by 0.5 V (first resistor), by 1 V across the cell, and by 0.5 V (second resistor):
V_B - V_A = -0.5 - 1 - 0.5 = -2 V.
…
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