Q.A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
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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. …
Concept: Internal Resistance – During charging, the battery’s terminal voltage is not its emf; it is the emf plus the voltage drop across its internal resistance because the charging current flows into the positive terminal.
Reasoning:
-
Net voltage in the circuit: supply voltage minus battery emf (which opposes charging).
Vnet=120 V−8.0 V=112 V.
-
Total resistance in series: Rext+r=15.5 Ω+0.5 Ω=16.0 Ω.
-
Charging current:
I=RtotalVnet=16.0112=7.0 A. …
During charging, the battery's terminal voltage is the sum of its emf and the voltage drop across its internal resistance (since current flows into the positive terminal). Here, the terminal voltage is 11.5 V, and the series resistor limits the charging current to a safe value.
1. Understanding the situation — charging a battery
When a battery is being charged, current is forced into its positive terminal. This is the opposite of discharging, where current flows out of the positive terminal.
The key difference:
- Discharging: terminal voltage V=ε−Ir (emf minus internal drop)
- Charging: terminal voltage V=ε+Ir (emf plus internal drop)
Why? Because the internal resistance r opposes the flow of current. During charging, the external supply pushes current against the battery's emf, so the internal resistance causes an additional voltage rise across the battery terminals.
A common mistake is to use the discharging formula V=ε−Ir for charging. Always check the direction of current relative to the battery's polarity.
2. Finding the charging current
The circuit: a 120 V DC supply, a series resistor R=15.5 Ω, and the battery (emf ε=8.0 V, internal resistance r=0.5 Ω) all in series.
The net voltage driving current is the supply voltage minus the battery's back emf (since the battery opposes the charging current):
Vnet=120 V−8.0 V=112 V
The total resistance in the circuit is:
Rtotal=R+r=15.5 Ω+0.5 Ω=16.0 Ω
So the charging current is:
I=RtotalVnet=16.0 Ω112 V=7.0 A
| Quantity | Value |
|----------|-------|
| Supply voltage | 120 V |
| Battery emf | 8.0 V |
| Net driving voltage | 112 V |
| Total resistance | 16.0 Ω |
| Charging current | 7.0 A |
3. Terminal voltage during charging
The terminal voltage of the battery is the voltage measured across its terminals. During charging:
Vterminal=ε+Ir
Substitute:
Vterminal=8.0 V+(7.0 A)(0.5 Ω)=8.0 V+3.5 V=11.5 V …
Method: Kirchhoff’s Voltage Law (KVL) for Charging Circuits
When a battery is being charged, current flows into its positive terminal — opposite to discharge. This means the internal resistance causes a voltage rise across it (opposing the charging current), so the terminal voltage becomes:
Vterminal=emf+Ir
Steps
-
Draw the circuit and assign current direction
- The 120 V supply pushes current through the series resistor (15.5 Ω) into the battery’s positive terminal.
- Current I flows clockwise: supply → series resistor → battery (positive to negative) → back to supply.
-
Apply KVL around the loop
Starting from the negative terminal of the supply, going clockwise:
+120 V−I(15.5 Ω)−I(0.5 Ω)−8.0 V=0
(The battery’s emf opposes the charging current, so it appears as a drop.)
- Solve for current I
120−8=I(15.5+0.5)
112=I×16
I=16112=7.0 A
- Find terminal voltage during charging Using the charging formula:
Vterminal=emf+Ir=8.0+(7.0×0.5)
Vterminal=8.0+3.5=11.5 V
--- …
Common Mistakes Students Make on This Internal Resistance Problem
Mistake 1: Using the Wrong Sign for Internal Resistance During Charging
The error: Students treat the internal resistance drop as Ir (subtracting it from emf), just like in a discharging circuit. They write:
V=ε−Ir
This is wrong for charging.
Why it's wrong: During charging, current flows into the battery's positive terminal. The internal resistance still opposes current flow, so the potential drop across r is +Ir (adds to the emf). The correct relation is:
V=ε+Ir
How to avoid: Always ask: Is the battery supplying energy (discharging) or receiving energy (charging)?
- Discharging: V=ε−Ir
- Charging: V=ε+Ir
Mistake 2: Forgetting to Find the Current First
The error: Students plug numbers directly into V=ε+Ir without calculating I from the full circuit.
Why it's wrong: The current I depends on the total resistance in the loop — the external resistor and the internal resistance.
Correct approach: Apply Kirchhoff's voltage law to the entire charging loop:
120 V−I(15.5 Ω)−I(0.5 Ω)−8.0 V=0
Solve for I:
120−8=I(15.5+0.5)
112=I×16
I=7.0 A
How to avoid: Draw the circuit. Label all resistances in series. Write one loop equation before touching the terminal voltage formula.
Mistake 3: Confusing Terminal Voltage with Emf
The error: Students answer V=8.0 V (the emf), thinking terminal voltage equals emf during charging.
Why it's wrong: Terminal voltage is the voltage across the battery terminals — which includes the Ir drop. During charging, this is greater than the emf.
Correct calculation:
V=ε+Ir=8.0+(7.0×0.5)=8.0+3.5=11.5 V
How to avoid: Remember: Terminal voltage is what a voltmeter reads across the battery. During charging, it's higher than emf; during discharging, it's lower.
Mistake 4: Giving a Vague or Incorrect Purpose for the Series Resistor
The error: Students say "to limit current" without explaining why limiting is necessary, or they say "to increase voltage." …
Showing the 12 most recent of 14 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.A battery supplies 0.6 A current when a 3 Ω resistor is connected with it. When the resistor 3 Ω is replaced by 6 Ω, the current is reduced to 0.4 A. Then, the internal resistance of the battery is (A) 3Ω (B) 9Ω (C) 6Ω (D) 12Ω (E) 2Ω
›Reveal solutionSolution
Equating the EMF for both cases gives internal resistance r=3 Ω.
The EMF of the battery equals I(R+r) in each case:
E=0.6(3+r)andE=0.4(6+r)
Setting them equal:
0.6(3+r)=0.4(6+r) …
- KEAM 2026Set eng-2026-04194 marksMCQQ.Two cells each of 2 V and internal resistance 0.1 Ω are connected in parallel combination. This combination is equivalent to a single cell with emf and internal resistance of (A) 1 V and 0.05 Ω (B) 2 V and 0.05 Ω (C) 2 V and 0.1 Ω (D) 4 V and 0.05 Ω (E) 4 V and 0.1 Ω
›Reveal solutionSolution
Identical cells in parallel keep the same emf and give internal resistance r/n.
For n identical cells (emf ε, internal resistance r) connected in parallel:
- Equivalent emf =ε (unchanged).
- Equivalent internal resistance =nr. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Two cells of emfs 2 V and 6 V each with internal resistance of 1 Ω are connected in series to an external resistance 8 Ω. Then the current in the circuit is (A) 2 A (B) 0.5 A (C) 1 A (D) 0.8 A (E) 1.5 A
›Reveal solutionSolution
Total emf 8 V across total resistance 10 Ω gives I=0.8 A.
Circuit. Two cells (2 V and 6 V) in series, aiding, so net emf =2+6=8 V. Total resistance = external 8 Ω + two internal 1 Ω eac …
- KEAM 2026Set eng-2026-04214 marksMCQQ.Three cells of 3V, 4V and 4V with respective internal resistances 0.5Ω, 0.75Ω and 0.75Ω are connected in series to a resistor of 4Ω. Then the current in the circuit is (A) 1A (B) 0.5A (C) 0.25A (D) 0.75A (E) 0.67A
›Reveal solutionSolution
Series loop: net driving EMF is 3V and total resistance 6Ω, giving 0.5A.
The internal resistances add to the external resistor in series:
Rtotal=0.5+0.75+0.75+4=6Ω.
The only combination of the three cell EMFs (3,4,4V) that gives a current matching the options is when the 3V cell opposes the two 4V cells (or two cells oppose), leaving a net EMF of 3V: …
- KEAM 2026Set eng-2026-04224 marksMCQQ.If three cells each of emf 2 V and internal resistance 1Ω are connected to a resistor of 4.5 Ω as shown, then the current through the resistor is (A) 1 A (B) 0.67 A (C) 0.33 A (D) 2 A (E) 0.5 A
›Reveal solutionSolution
The two side-by-side cells are in parallel (2 V, 0.5Ω); this combines in series with the third cell, then apply Ohm's law across the 4.5Ω resistor.
Two identical 2 V, 1Ω cells in parallel give an emf of 2 V and internal resistance 21(1)=0.5Ω. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.When ′n′ identical cells are connected in parallel, (A) net voltage increases (B) net current increases (C) net voltage decreases (D) net current decreases (E) total internal resistance increases
›Reveal solutionSolution
Connecting identical cells in parallel keeps the voltage the same but increases the current the battery can deliver.
Concept and Intuition
Cells in parallel all share the same EMF, so the net terminal voltage stays equal to that of a single cell. However, the effective internal resistance drops to r/n, allowing a larger current to be supplied to the load.
Step-by-Step Solution
- EMF of parallel combination = EMF of one cell (voltage unchanged).
- Internal resistance = r/n (decreases). …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The maximum current drawn from a battery of 12 V with internal resistance of 0.5 Ω is (A) 16 A (B) 12 A (C) 24 A (D) 30 A (E) 4 A
›Reveal solutionSolution
The largest current a source can deliver is limited only by its own internal resistance (external R=0), so Imax=E/r.
The terminal current of a cell is I=R+rE. This is maximised when the external resistance R→0 (a short circuit), giving …
- KEAM 2025Set eng-2025-04274 marksMCQQ.If n identical cells each of emf E and internal resistance r are connected in parallel, the total EMF and total internal resistance of the combination, respectively, are (A) nE, nr (B) E, nr (C) E, nr (D) nE, 2nr (E) nE, nr
›Reveal solutionSolution
Cells in parallel keep the same EMF (E) but their internal resistances combine like parallel resistors, so the net internal resistance is nr.
When n identical cells, each of EMF E and internal resistance r, are joined in parallel, all their positive terminals are tied together and all negatives together. Since every cell drives the same potential difference, the combined EMF is just E (not nE — that would be a series stack).
The n internal resistances r are now in parallel: …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.An electric cell does 10 J of work in carrying a charge of 5 C around a simple closed circuit. The electromotive force of the cell is (A) 0.5 V (B) 1.5 V (C) 1 V (D) 6 V (E) 2 V
›Reveal solutionSolution
EMF is work done per unit charge: ε=W/q=10/5=2 V.
The electromotive force is defined as the work done by the cell per unit charge carried around the circuit: …
- KEAM 2024Set eng-2024-06064 marksMCQQ.If a cell of 12 V emf delivers 2 A current in a circuit having a resistance of 5.8 Ω, then the internal resistance of the cell is (A) 1 Ω (B) 0.2 Ω (C) 0.3 Ω (D) 0.6 Ω (E) 0.8 Ω
›Reveal solutionSolution
The emf drives current through external plus internal resistance: ε=I(R+r).
Applying ε=I(R+r):
12=2(5.8+r)⇒6=5.8+r⇒r=0.2 Ω. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.The terminal potential difference of a cell in the open circuit is 2 V. When the cell is connected to a 10Ω resistor, the terminal potential difference falls to 1.5 V. The internal resistance of the cell is (A) 310Ω (B) 910Ω (C) 720Ω (D) 615Ω (E) 213Ω
›Reveal solutionSolution
Open-circuit voltage = EMF = 2 V; loaded current 0.15 A; r=(E−V)/I=0.5/0.15=10/3Ω.
In the open circuit the terminal voltage equals the EMF: E=2 V.
When connected across R=10 Ω, terminal voltage V=1.5 V, so the current is
I=RV=101.5=0.15 A. …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The y-intercept of the graph between the terminal voltage V with load resistance R along y and x - axis, respectively, of a cell with internal resistance r, as shown, is (A) ε (B) −ε (C) Rε (D) εR (E) −εR
›Reveal solutionSolution
The straight-line terminal-voltage plot meets the V-axis at the cell's emf, so the y-intercept =ε.
For a cell of emf ε and internal resistance r driving a load R,
V=ε−Ir,I=R+rε. …
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