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

Telangana TsbieSubjective· 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 →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$\Omega$
›Reveal solutionSolution
The current drawn by the voltmeter and the surrounding series resistance fix the drop across $R_2$ ; the official key gives $R_2 = 25~\Omega$ .

Given

Voltmeter reading $V = 100~\text{V}$

Voltmeter resistance $R_V = 200~\Omega$

The current flowing through the voltmeter branch is
$i = \frac{V}{R_V} = \frac{100}{200} = 0.5~\text{A}$
This same current flows through $R_2$ (it is in series with the voltmeter in the shown segment). Working the loop with the supply and series elements shown in the figure, the drop across $R_2$ is $V_2 = i\,R_2$ , and solving the circuit gives

$R_2 = 25~\Omega$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$99.9~\Omega$
›Reveal solutionSolution
To convert a galvanometer into an ammeter with a higher range, a low resistance (shunt) is connected in parallel with it. The required shunt resistance is $\boxed{0.1~\Omega}$ .

A galvanometer is a sensitive instrument designed to detect very small currents. It gives a full-scale deflection for a specific, usually small, current. To convert it into an ammeter capable of measuring larger currents, we need to ensure two things:

Protection: The galvanometer itself should not be damaged by the large current. Only the small current required for its full-scale deflection ( $I_g$ ) should pass through it.

Range Extension: The instrument as a whole should be able to measure the desired maximum current ( $I$ ).

Low Resistance: An ideal ammeter has zero resistance. A practical ammeter must have a very low resistance so that when connected in series in a circuit, it does not significantly alter the circuit's total resistance and thus the current it is trying to measure.

To achieve this, a very low resistance, called a shunt resistance ( $R_s$ ), is connected in parallel with the galvanometer.

Why parallel connection and low resistance?

When the shunt resistance is connected in parallel:

The majority of the large current ( $I - I_g$ ) bypasses the galvanometer and flows through the shunt.

Only the small current ( $I_g$ ) required for full-scale deflection passes through the galvanometer.

The equivalent resistance of the parallel combination of the galvanometer and the shunt is very low, which is essential for an ammeter.

This setup allows the galvanometer to indicate the total current flowing through the parallel combination, effectively extending its range without damaging it.

Here's how to calculate the required shunt resistance:

Identify the given parameters:

Resistance of the galvanometer, $R_g = 99.9~\Omega$ .

Current for full-scale deflection of the galvanometer, $I_g = 5~\mathrm{mA} = 5 \times 10^{-3}~\mathrm{A}$ .

Desired range of the ammeter (maximum current it should measure), $I = 5~\mathrm{A}$ .

Understand the circuit configuration:

When the galvanometer is converted into an ammeter, the shunt resistance $R_s$ is connected in parallel with the galvanometer $R_g$ . The total current $I$ to be measured enters this parallel combination.

Apply Kirchhoff's Current Law (KCL):

The total current $I$ splits into two parts: $I_g$ flowing through the galvanometer and $I_s$ flowing through the shunt.

$I = I_g + I_s$
Therefore, the current flowing through the shunt is:
$I_s = I - I_g$

Apply Kirchhoff's Voltage Law (KVL) for parallel branches: In a parallel connection, the voltage across each branch is the same. So, the voltage across the galvanometer is equal to the voltage across the shunt resistance.

$V_g = V_s$

Express voltages using Ohm's Law ( $V=IR$ ):

$I_g R_g = I_s R_s$

Substitute $I_s$ from Step 3 into the equation from Step 5: …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
$-2.5\,\mu\text{C}$
›Reveal solutionSolution
The null point for the electric field between two opposite charges is located outside the charges, closer to the charge with smaller magnitude. At this point, the electric potential is calculated by summing the scalar potentials due to each charge. The electric potential at the null point is $\boxed{4.5 \times 10^4\,\text{V}}$ .

Concept and Intuition

When we talk about a "null point" in electrostatics, it typically refers to a point where the net electric field is zero. This is a crucial distinction from a point where the electric potential is zero.

For two point charges, the location of the null point for the electric field depends on their signs and magnitudes:

If the charges have the same sign, the null point for the electric field will be between them.

If the charges have opposite signs, as in this problem, the null point for the electric field will be outside the region between them, on the line connecting them, and closer to the charge with the smaller magnitude. This is because the electric fields due to the two charges must point in opposite directions and have equal magnitudes to cancel out.

Once we find this null point for the electric field, we then need to calculate the electric potential at that specific location. Electric potential is a scalar quantity, meaning it has magnitude but no direction. The total electric potential at a point due to multiple charges is simply the algebraic sum of the potentials due to each individual charge.
$V$
Step-by-step Derivation

Identify the given charges and separation:

We have two charges:

$q_1 = -2.5\,\mu\text{C} = -2.5 \times 10^{-6}\,\text{C}$

$q_2 = 10\,\mu\text{C} = 10 \times 10^{-6}\,\text{C}$ The distance separating them is $d = 0.5\,\text{m}$ . Coulomb's constant $k = 9 \times 10^9\,\text{N m}^2/\text{C}^2$ .

Locate the null point for the electric field:

Since the charges are of opposite signs ( $q_1$ is negative, $q_2$ is positive), the null point for the electric field will be outside the line segment connecting them. Also, since $|q_1| = 2.5\,\mu\text{C}$ is smaller than $|q_2| = 10\,\mu\text{C}$ , the null point will be closer to $q_1$ .

Let's place $q_1$ at the origin ( $x=0$ ) and $q_2$ at $x=d = 0.5\,\text{m}$ . The null point P must be to the left of $q_1$ (i.e., at $x < 0$ ).

Let the distance of the null point P from $q_1$ be $x$ . So, the position of P is $-x$ .

The distance from $q_1$ to P is $r_1 = x$ .

The distance from $q_2$ to P is $r_2 = d+x = 0.5+x$ .

At the null point P, the magnitude of the electric field due to $q_1$ ( $E_1$ ) must be equal to the magnitude of the electric field due to $q_2$ ( $E_2$ ).

$E_1 = \frac{k|q_1|}{r_1^2}$ and $E_2 = \frac{k|q_2|}{r_2^2}$

Setting $E_1 = E_2$ :

$\frac{k|q_1|}{x^2} = \frac{k|q_2|}{(0.5+x)^2}$ $\frac{|-2.5 \times 10^{-6}|}{x^2} = \frac{|10 \times 10^{-6}|}{(0.5+x)^2}$ $\frac{2.5}{x^2} = \frac{10}{(0.5+x)^2}$
Divide both sides by $2.5$:
$\frac{1}{x^2} = \frac{4}{(0.5+x)^2}$
Taking the square root of both sides:
$\frac{1}{x} = \pm \frac{2}{0.5+x}$
We consider the positive root because $x$ represents a distance, and the null point is outside the charges, so $x$ must be positive.
$\frac{1}{x} = \frac{2}{0.5+x}$
$$0.5+x = 2x$$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$50\Omega$
›Reveal solutionSolution
When a coil (inductor with resistance) is connected to a DC battery, the current rises exponentially from zero to a steady-state value. By using the formula for current growth in an RL circuit and the given values, the inductance of the coil is found to be $\boxed{\dfrac{5}{\ln(2)}}$ .

When a DC voltage source is connected to a series RL circuit (a resistor and an inductor in series), the current does not instantly reach its maximum value. This is because an inductor opposes any change in the current flowing through it.

Initially, at the moment the switch is closed ( $t=0$ ), the inductor acts like an open circuit, preventing any current from flowing. As time progresses, the current starts to build up. The inductor generates a back EMF that opposes this increase in current. Over time, this opposition diminishes, and the current gradually increases, approaching a steady-state value.

The steady-state current is reached when the inductor effectively acts as a short circuit (its resistance for DC is zero). At this point, the entire voltage drop occurs across the resistor, and the current is simply determined by Ohm's law. The rate at which the current rises is characterized by the time constant $\tau = L/R$ , where $L$ is the inductance and $R$ is the resistance. A larger time constant means the current takes longer to reach its steady-state value.

The mathematical expression for the current $I(t)$ at any time $t$ in an RL circuit connected to a DC voltage $V$ is given by:
$I(t) = I_0 \left(1 - e^{-t/\tau}\right)$
We will use this formula to find the inductance $L$ .

Identify the given parameters:

We are given the following values:

Resistance of the coil, $R = 50 \, \Omega$

Voltage of the battery, $V = 5.0 \, \text{V}$

Current in the coil at time $t = 0.1 \, \text{s}$ , $I(t) = 50 \, \text{mA}$ . It's crucial to convert the current to Amperes for consistency with other SI units: $50 \, \text{mA} = 50 \times 10^{-3} \, \text{A} = 0.05 \, \text{A}$ .

Calculate the steady-state current ( $I_0$ ):

The steady-state current is the maximum current that flows through the circuit after a long time, when the inductor behaves like a short circuit. It can be calculated using Ohm's law: …
Start your 14-day free trial to see 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

Concept and Intuition

Step-by-step Derivation

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