Power $P$ is to be delivered to a device via transmission cables having resistance $R_C$. If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.

Power P is to be delivered to a device via transmission cables having resistance RC. If V is the voltage acros

NCERT Exemplar · Q14

P

Rajasthan RbseSubjective· 2mImportance★★★★★

71% · 30/42 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

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 $R_{\text{int}}$ ).

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 ( $\Omega$ ).

For a cell or battery, the relationship between the emf ( $\mathcal{E}$ ), the terminal voltage ( $V$ ), the current ( $I$ ), and the internal resistance ( $r$ ) is given by:

V = \mathcal{E} - I r

This is the single most important equation to remember.

V = \mathcal{E} - I r

Here:

$\mathcal{E}$ 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 $I r$ 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)	$I = 0$	$V = \mathcal{E}$	No current means no internal drop.
Small load (e.g., a dim bulb)	Small $I$	$V \approx \mathcal{E}$	The $I r$ term is tiny.
Large load (e.g., a powerful motor)	Large $I$	$V < \mathcal{E}$ significantly	The $I r$ drop becomes noticeable.
Short circuit (wires directly across terminals)	Very large $I$	$V \approx 0$	Almost all voltage is lost inside the battery; the battery heats up and may be damaged.

Watch out

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 $0.5\ \Omega$ ). 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 $P_{\text{loss}} = I^2 r$ . 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 ( $R_{\text{load}} = r$ ). 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 ( $\mathcal{E}$ ) — 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 $\mathcal{E}$ .
The internal resistor $r$ drops some voltage: $V_{\text{drop}} = I r$ (Ohm's law).
The voltage available at the terminals ( $V$ ) is what's left:

V = \mathcal{E} - I r

Why this makes sense:

If $I = 0$ (open circuit), $V = \mathcal{E}$ — you measure the full EMF.
If $I$ increases, $V$ decreases — the internal drop grows.
If $I$ is huge (short circuit), $V \to 0$ 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: $I_{\text{short}} = \frac{\mathcal{E}}{r}$

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: $I = \frac{\mathcal{E}}{R + r}$
Power delivered to the external load ( $P_{\text{out}}$ ):

P_{\text{out}} = I^2 R = \left( \frac{\mathcal{E}}{R + r} \right)^2 R

Why this matters:

Power is not simply $\mathcal{E}^2 / R$ — because $r$ limits current. …

Start your 14-day free trial to unlock the full solution →

The power wasted in the transmission cables is $P_{\text{waste}} = I^2 R_C$ , and it can be reduced by increasing the transmission voltage $V$ (which lowers the current $I$ for the same delivered power $P$ ).

The Core Idea: Why Wasted Power Depends on Current, Not Voltage

When you deliver power to a device, the transmission cables themselves have resistance $R_C$ . Any current flowing through them must obey Ohm’s law — and that means some voltage drops across the cables, and power is dissipated as heat. This is unavoidable but minimizable.

The key insight: the power wasted in the cables is $I^2 R_C$ , not $V^2 / R_C$ . Why? Because the voltage $V$ in the problem is the voltage across the device (the load), not across the cables. The cables and the load are in series, so the same current $I$ flows through both. The wasted power is purely resistive loss in the cables.

Watch out

A common mistake is to write $P_{\text{waste}} = V^2 / R_C$ . That would be true only if $V$ were the voltage across the cables themselves — but here $V$ is the load voltage. The cable voltage drop is $I R_C$ , which is much smaller than $V$ in an efficient system.

Step-by-Step Derivation

1. Identify the given quantities.

We have:

$R_C$ = resistance of the transmission cables (fixed, determined by wire material and length).
$V$ = voltage across the load (the device).
$I$ = current through the load (and also through the cables, since they are in series).

2. Write the power delivered to the device.

The useful power is:

P = V I

3. Find the power wasted in the cables.

The cables have resistance $R_C$ and carry current $I$ . The power dissipated as heat in the cables is:

P_{\text{waste}} = I^2 R_C

This is the Joule heating loss. It does not depend on $V$ directly — only on $I$ and $R_C$ .

P_{\text{waste}} = I^2 R_C

4. How can this waste be reduced?

Since $R_C$ is fixed by the cable material and length (you can’t easily change it after installation), the only handle is to reduce the current $I$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ13 — Why are alloys used for making standard resistance coils?→Next questionQ15 — A cell of emf E and internal resistance r is connected across an exter…

Q.Power $P$ is to be delivered to a device via transmission cables having resistance $R_C$ . If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.

Concept understanding — Internal Resistance

Internal Resistance – The Battery That Fights Itself

The Intuition

The Precise Statement

What Happens in Different Situations?

Why Does Internal Resistance Matter?

Why this formula?

Internal Resistance: Why the Key Formulas Hold

1. The Core Idea: A Real Battery = An Ideal Source + A Resistor

2. The Terminal Voltage Formula

3. The Short-Circuit Current Formula

4. Power Delivered to External Load

The Core Idea: Why Wasted Power Depends on Current, Not Voltage

Step-by-Step Derivation

Method: Minimizing Resistive (Joule) Power Loss in a Series Element

Steps

Q.Power PPP is to be delivered to a device via transmission cables having resistance RCR_CRC​. If VVV is the voltage across RRR and III the current through it, find the power wasted and how can it be reduced.

The Core Idea: Why Wasted Power Depends on Current, Not Voltage

Step-by-Step Derivation

🎓Unlock everything free for 14 days

Q.Power $P$ is to be delivered to a device via transmission cables having resistance $R_C$ . If $V$ is the voltage across $R$ and $I$ the current through it, find the power wasted and how can it be reduced.

Unlock everything free for 14 days