Case Study : The following figure shows a circuit diagram. We can find the currents through and potential diff

Question

Figure — Figure — CBSE 2023 55/1/1 Q34 — FigureFigure — CBSE 2023 55/1/1 Q34

Q. Case Study : The following figure shows a circuit diagram. We can find the currents through and potential differences across the different resistors using Kirchhoff's rules. Answer the following questions based on the above :

(a) Which points are at the same potential in the circuit?

(b) What is the current through arm bg?

R_3

(OR)

R_2

CBSECBSE Class XII Board 2023Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Wheatstone Bridge Symmetry

Wheatstone Bridge Symmetry

Four resistors are arranged in a diamond. A battery sits across the top and bottom points; a sensitive galvanometer bridges the left and right points. The question is: when does no current flow through the galvanometer? The answer is symmetry.

The Intuition

Treat the bridge as two parallel voltage dividers — two resistors in series on the left, two in series on the right — with the galvanometer joining their midpoints. If both dividers produce the same midpoint voltage, there is no potential difference across the galvanometer and no current flows: the bridge is balanced. This happens when the resistance ratio on the left arm equals the ratio on the right arm.

Note

Symmetry here means equal ratios, not equal resistances. $1\ \Omega$ with $2\ \Omega$ on one side balances $100\ \Omega$ with $200\ \Omega$ on the other.

The Precise Statement

Label the four resistors: $R_1$ top-left, $R_2$ bottom-left, $R_3$ top-right, $R_4$ bottom-right. The battery connects the top junction (between $R_1$ , $R_3$ ) to the bottom junction (between $R_2$ , $R_4$ ); the galvanometer connects the two midpoints. The bridge is balanced when

\frac{R_1}{R_2} = \frac{R_3}{R_4}

Important

The balance condition is independent of the battery voltage and of the galvanometer's resistance — it depends only on the four resistor values.

Why This Matters

Unknown resistance: put the unknown in place of $R_4$ , adjust the others until balanced, then $R_4 = R_3\,\dfrac{R_2}{R_1}$ .
Strain gauges: stretching slightly changes a resistance, unbalancing the bridge; the deflection measures the strain.
Temperature sensors: a temperature-dependent resistor unbalances the bridge in proportion to the change. …

Why this formula?

Wheatstone Bridge Symmetry — Why the Formula Holds

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge. The key result is:

When the bridge is balanced, no current flows through the galvanometer, and:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Let's understand why this is true — not just memorize it.

1. The Circuit Setup

A Wheatstone bridge has four resistors arranged in a diamond:

$R_1$ and $R_2$ in series on the left branch
$R_3$ and $R_4$ in series on the right branch
A galvanometer (sensitive current detector) connects the midpoints of the two branches
A battery connects across the top and bottom

        A
       / \
      /   \
     R1    R3
     |     |
     G-----|   (G = galvanometer)
     |     |
     R2    R4
      \   /
       \ /
        B

2. The Condition for Balance — "No Current Through G"

Balance means the galvanometer shows zero deflection — no current flows through it. This implies points C (between $R_1$ and $R_2$ ) and D (between $R_3$ and $R_4$ ) are at the same electric potential. If $V_C = V_D$ , no potential difference exists across the galvanometer, so no current flows.

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

With no current through G, the left branch carries a single current $I_1$ and the right branch a single current $I_2$ .

Step 2: Equal potentials at the midpoints

From A (common top) to the midpoints, since $V_C = V_D$ :

I_1 R_1 = I_2 R_3 \quad \text{(1)}

From the midpoints to B (common bottom), since $V_C = V_D$ :

I_1 R_2 = I_2 R_4 \quad \text{(2)}

Step 3: Divide (1) by (2)

\frac{I_1 R_1}{I_1 R_2} = \frac{I_2 R_3}{I_2 R_4}

The currents $I_1$ and $I_2$ cancel: …

Part (b)Concept understanding — Power Dissipation in Resistors

Power Dissipation in Resistors

Whenever charge is driven through a resistor, electrical energy is converted into heat. The rate of this conversion is the power dissipated.

Why a Resistor Heats Up

Inside a resistor, drifting electrons repeatedly collide with the vibrating lattice ions. Each collision transfers kinetic energy to the lattice, raising its temperature. The source (battery or AC supply) continually does work to keep the current flowing, and that work reappears as heat. This is Joule heating.

The Power Formulas (DC)

The power delivered to any device carrying current $I$ across a potential difference $V$ is

P = VI

For an ohmic resistor $V = IR$ , so this can be written in three equivalent forms:

P = VI = I^2 R = \frac{V^2}{R}

The SI unit is the watt (W), where $1\ \text{W} = 1\ \text{J s}^{-1}$ .

Which form to use depends on what is fixed:

Series elements share the same current, so $P = I^2 R$ shows the larger resistor dissipates more.
Parallel elements share the same voltage, so $P = V^2/R$ shows the smaller resistor dissipates more.

The total heat produced in time $t$ is $Q = Pt = I^2 R t$ — Joule's law of heating.

Power Dissipation with AC

With alternating current the instantaneous power $p(t) = i^2(t)R$ fluctuates, but a resistor still only dissipates energy (it never returns any). The average power over a cycle is written with root-mean-square values:

P_{\text{avg}} = I_{\text{rms}}^2 R = \frac{V_{\text{rms}}^2}{R} = V_{\text{rms}} I_{\text{rms}}

where for a sinusoid $I_{\text{rms}} = I_m/\sqrt{2}$ and $V_{\text{rms}} = V_m/\sqrt{2}$ . This is precisely why rms values are defined: an AC of rms value $I_{\text{rms}}$ heats a resistor at the same average rate as a steady DC of value $I_{\text{rms}}$ .

Important

A pure resistor has power factor 1 — voltage and current are in phase, so all the power supplied is dissipated. In inductors and capacitors, by contrast, the average dissipated power is zero; energy is only stored and returned.

Worked Example

A $100\ \Omega$ resistor carries a current of $0.5\ \text{A}$ . …

Why this formula?

Power Dissipation in Resistors — Why the Formula Holds

Let's build this from first principles. The goal is to understand why a resistor dissipates power as heat, and how the formula $P = I^2 R$ (and its equivalents) arise naturally.

1. What is "Power" in an Electrical Circuit?

Power is the rate of energy transfer — how much energy is converted from one form to another per unit time.

In a resistor, electrical energy is converted into heat (thermal energy).
The fundamental definition of electrical power is:

P = V \cdot I

where:

$P$ = power (watts, W)
$V$ = voltage across the component (volts, V)
$I$ = current through the component (amperes, A)

Why this definition?

Voltage is energy per unit charge ( $V = \frac{W}{q}$ ), and current is charge per unit time ( $I = \frac{q}{t}$ ). Multiplying them gives energy per unit time — exactly power.

2. How Does a Resistor Behave? — Ohm's Law

A resistor obeys Ohm's Law:

V = I \cdot R

where $R$ is resistance (ohms, $\Omega$ ). This is an empirical law — it describes how real resistors behave: the voltage across them is proportional to the current through them.

3. Deriving the Power Dissipation Formulas

We start with $P = V I$ and substitute Ohm's Law in two ways.

Case A: Express power in terms of $I$ and $R$

Replace $V$ with $I R$ :

P = (I R) \cdot I = I^2 R

Interpretation:

For a fixed resistance, power grows with the square of current.
Doubling current quadruples the heat generated — this is why high currents cause wires to overheat.

Case B: Express power in terms of $V$ and $R$

Replace $I$ with $\frac{V}{R}$ :

P = V \cdot \left( \frac{V}{R} \right) = \frac{V^2}{R}

Interpretation:

For a fixed voltage, power is inversely proportional to resistance.
A low-resistance resistor (like a short circuit) dissipates huge power at a given voltage — that's why short circuits are dangerous.

4. The Physical "Why" — Energy Conversion at the Atomic Level

Why does this energy turn into heat?

Electrons moving through a resistor collide with the atoms of the material.
Each collision transfers kinetic energy from the electron to the atom, making the atom vibrate more — i.e., heating up the resistor.
The rate at which this energy is lost by the electrons (and gained by the lattice) is exactly $P = I^2 R$ .

Key insight:

The $I^2$ term appears because:

More current = more electrons per second. …

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The two parallel branches are each $10\ \Omega$ , so the $1.0$ A main current splits equally ( $0.5$ A). (a) {a,b,c}, {f,g,h}, {d,e} are equipotential; $I_{bg}=0.5$ A; $V_{R_3}=2.5$ V. (c) $P_{R_2}=1.25$ W.

(Values read from the figure: $\mathcal{E}=6.0$ V, $r=1\ \Omega$ , $R_1=10\ \Omega$ in arm bg, and $R_2=R_3=5\ \Omega$ in series in the right branch.)

Setting up the circuit

Points joined by plain (resistanceless) wires are at the same potential. The right branch has $R_2+R_3=5+5=10\ \Omega$ ; the left branch is $R_1=10\ \Omega$ . These two equal $10\ \Omega$ branches are in parallel:

R_\parallel=\frac{10\times10}{10+10}=5\ \Omega,\qquad R_{tot}=r+R_\parallel=1+5=6\ \Omega.

Main current from the cell:

I=\frac{\mathcal{E}}{R_{tot}}=\frac{6.0}{6}=1.0\ \text{A}.

Because the two branches have equal resistance, the current divides equally: $0.5$ A in each.

Part (a)

(a) Same-potential points. Along the top rail a–b–c and the bottom rail f–g–h the connecting wires have no resistance, so each rail is an equipotential; the wire joining d and e (between $R_2$ and $R_3$ ) puts them at a common (intermediate) potential. …

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

CBSE 2026Set 55/1/11 markMCQ
$(P_1, V)$
›Reveal solutionSolution
Each heater's resistance is found from its rated power and voltage; in series across $\frac{V}{2}$ , the total power dissipated is $\frac{P_1 P_2}{4(P_1 + P_2)}$ .

Why this approach works

When a device is rated at $(P, V)$ , it means that at voltage $V$ it consumes power $P$ . This rating tells us the device's resistance through $P = \frac{V^2}{R}$ , so $R = \frac{V^2}{P}$ . Once we know the resistances, we can treat the heaters as ordinary resistors in a series circuit and calculate the actual power consumed at the new operating voltage.

The key insight: rated values describe behavior at a specific voltage, but resistance is an intrinsic property that doesn't change. We extract the resistance from the rating, then analyze the actual circuit.

Step-by-step solution

Find the resistance of each heater from its rating.

For heater 1 rated at $(P_1, V)$ :

$R_1 = \frac{V^2}{P_1}$
For heater 2 rated at $(P_2, V)$ :
$R_2 = \frac{V^2}{P_2}$

Calculate the total resistance in series.

When connected in series, resistances add:

$R_{\text{total}} = R_1 + R_2 = \frac{V^2}{P_1} + \frac{V^2}{P_2} = V^2 \left(\frac{1}{P_1} + \frac{1}{P_2}\right) = V^2 \cdot \frac{P_1 + P_2}{P_1 P_2}$

Apply the actual supply voltage.

The combination is connected across $\frac{V}{2}$ . The power consumed by a resistor is:

$P = \frac{V_{\text{applied}}^2}{R_{\text{total}}}$
Substituting:
$P = \frac{\left(\frac{V}{2}\right)^2}{V^2 \cdot \frac{P_1 + P_2}{P_1 P_2}} = \frac{\frac{V^2}{4}}{V^2 \cdot \frac{P_1 + P_2}{P_1 P_2}}$

Simplify the expression. …
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CBSE 2026Set 55/2/11 markMCQ
$P_1$
›Reveal solutionSolution
The key idea is that in series, the combined resistance is larger than either heater's resistance, so the total power drawn from the source is smaller. The assertion is true; the reason is false because power is inversely proportional to resistance for a fixed voltage, not proportional.

Concept and intuition

When you connect a device to a fixed DC voltage source $V$ , the power it consumes is given by $P = V^2 / R$ . For a fixed voltage, power is inversely proportional to resistance — a higher resistance draws less current and therefore consumes less power. The reason statement gets this backwards.

Now, when two heaters are joined in series, their resistances add up. Since each heater's resistance is $R_i = V^2 / P_i$ (from $P_i = V^2 / R_i$ ), the series combination has a total resistance $R_{\text{series}} = R_1 + R_2$ , which is larger than either $R_1$ or $R_2$ alone. With a larger resistance, the power drawn from the same voltage source must be smaller than the power drawn by either individual heater. In particular, it will be less than $P_1$ (the smaller power heater, which has the larger resistance). So the assertion is correct, but for a reason opposite to what is stated.

Step-by-step reasoning

Express each heater's resistance in terms of its rated power. For a heater rated at power $P$ when connected to voltage $V$ , we have $P = V^2 / R$ , so $R = V^2 / P$ . Therefore:

$R_1 = \frac{V^2}{P_1}, \quad R_2 = \frac{V^2}{P_2}.$
Since $P_2 > P_1$ , it follows that $R_2 < R_1$ (higher power means lower resistance).

Find the total resistance when they are in series.

$R_{\text{series}} = R_1 + R_2 = V^2 \left( \frac{1}{P_1} + \frac{1}{P_2} \right).$
Clearly $R_{\text{series}} > R_1$ (and also $> R_2$ ).

Compute the power consumed by the series combination. Using $P = V^2 / R$ again:

$P_{\text{series}} = \frac{V^2}{R_{\text{series}}} = \frac{V^2}{V^2 \left( \frac{1}{P_1} + \frac{1}{P_2} \right)} = \frac{1}{\frac{1}{P_1} + \frac{1}{P_2}} = \frac{P_1 P_2}{P_1 + P_2}.$

Compare $P_{\text{series}}$ with $P_1$ . Since $P_1 > 0$ , we have $P_1 + P_2 > P_2$ , so $P_{\text{series}} = \frac{P_1 P_2}{P_1 + P_2} < \frac{P_1 P_2}{P_2} = P_1.$ …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. If R1 and R2 are respectively the filament resistance of a 200 W bulb and a 100 W bulb designed to operate on the same voltage, then -$
(a) R1 = 2R2
(b) R2 = 2R1
(c) R2 = 4R1
(d) R1 = 4R2
›Reveal solutionSolution

Since power $P = V^2/R$ at fixed voltage, the lower-power bulb has the higher filament resistance.

Both bulbs operate at the same voltage $V$ . Using $P = \dfrac{V^2}{R}$ , so $R = \dfrac{V^2}{P}$ .

For the 200 W bulb: $R_1 = \dfrac{V^2}{200}$

For the 100 W bulb: $R_2 = \dfrac{V^2}{100} = \dfrac{2V^2}{200} = 2R_1$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Three resistances 2 ohm, 3 ohm and 4 ohm are connected in parallel. The ratio of currents passing through them when a potential difference is applied across its ends will be$
(i) 6 : 4 : 3
(ii) 4 : 3 : 2
(iii) 6 : 3 : 2
(iv) 5 : 4 : 3
›Reveal solutionSolution

Same voltage across parallel branches gives currents in ratio 1/R, i.e. 6 : 4 : 3.

In a parallel combination the potential difference V is the same across each resistor, so the current through each is $I = V/R$ , inversely proportional to R. Thus …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Four resistances of 100 ohm each are connected in the form of a square. Then the effective resistance between any two diagonally opposite points is$
(i) 200 ohm
(ii) 400 ohm
(iii) 100 ohm
(iv) 150 ohm
›Reveal solutionSolution

Two series pairs (each 200 ohm) in parallel give 100 ohm across a diagonal.

Label the corners of the square P, Q, R, S with a 100 ohm resistor on each side. Take the diagonal P and R. There are two paths from P to R: P-Q-R (100 + 100 = 200 ohm) and P-S-R (100 + 100 = 200 ohm). These two pat …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In a Wheatstone bridge, if the battery and galvanometer are interchanged, then the deflection in galvanometer will$
(a) Change in previous direction
(b) Change in opposite direction
(c) Not change
(d) None of these
›Reveal solutionSolution

The Wheatstone bridge balance condition is symmetric in the battery and galvanometer arms (reciprocity theorem), so swapping them does not change the galvanometer's response.

A Wheatstone bridge has four resistances P, Q, R, S arranged in a diamond, with a battery across one diagonal and a galvanometer across the other. The balance condition is

$\dfrac{P}{Q} = \dfrac{R}{S}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Energy dissipated in LCR circuit is in$
(a) L only
(b) C only
(c) R only
(d) All of these
›Reveal solutionSolution

Over a full AC cycle, a pure inductor and a pure capacitor store and release energy with zero net dissipation; only the resistive element genuinely converts electrical energy to heat.

In an LCR series circuit driven by an AC source, the current and voltage across L and C are 90 degrees out of phase with each other, so the average power delivered to a pure inductor or a pure capacitor over one complete cycle is zero:

$P_L = P_C = 0 \quad (\text{average, over one cycle})$

…

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CBSE 2023Set 55/4/11 markMCQ
$200\ \text{W}$
›Reveal solutionSolution
In a series circuit, the bulb with the lowest rated power has the highest resistance, and since power dissipated in series is $P = I^2 R$ , the $50\ \text{W}$ bulb consumes the most power. The reason correctly states that current is same but voltage differs, but it does not explain why the $50\ \text{W}$ bulb gets maximum power — that requires linking resistance to rated power. So both statements are true, but Reason is not the correct explanation.

The Concept — Why This Works

The trap here is intuitive: we usually think a 200 W bulb is "more powerful." But that's when each bulb is connected individually to the same voltage (say 220 V). In that case, a higher wattage means it draws more current and glows brighter.

In series, the situation flips. The key idea:

Each bulb is designed for a fixed voltage (the mains voltage). Its resistance is fixed by $R = V^2 / P_{\text{rated}}$ .

A lower rated power means a higher resistance (since $P$ is in the denominator).

In series, current $I$ is the same through all bulbs. Power dissipated in a bulb is $P_{\text{actual}} = I^2 R$ .

So the bulb with the largest resistance (the 50 W bulb) dissipates the most power in series.

That's the core physics. Now let's check the statements carefully.

Step-by-Step Verification

1. Find the resistances of the bulbs.

Assume each bulb is rated for the same voltage $V$ (typically 220 V in household circuits, but the exact value doesn't matter — it cancels out).

Using $P = V^2 / R$ , we get $R = V^2 / P$ .

For 200 W bulb: $R_{200} = V^2 / 200$

For 100 W bulb: $R_{100} = V^2 / 100$

For 50 W bulb: $R_{50} = V^2 / 50$

Clearly, $R_{50} > R_{100} > R_{200}$ .

2. Connect them in series to the same source voltage $V$ .

Total resistance: $R_{\text{total}} = R_{200} + R_{100} + R_{50}$ .

Current in the circuit:
$I = \frac{V}{R_{\text{total}}}$
This current is the same through each bulb (series property).

3. Power consumed by each bulb in series.

For any bulb: $P_{\text{actual}} = I^2 R$ .

Since $I$ is common, the bulb with the largest $R$ gets the largest $P_{\text{actual}}$ .

That's the 50 W bulb. So Assertion (A) is true.

4. Check Reason (R).

Reason says: "In a series circuit, current is the same through each bulb, but the potential difference across each bulb is different."

This is a true statement about series circuits. …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. Of the two bulbs in a house, one glows brighter than the other. Which of the two has a larger resistance?$
(a) The brighter bulb
(b) The dim bulb
(c) Both have same resistance
(d) The brightness does not depend upon the resistance
›Reveal solutionSolution

Both bulbs share the same house-supply voltage, so power (and hence brightness) is inversely proportional to resistance: $P=V^2/R$ .

In a house, bulbs are connected in parallel across the same mains voltage $V$ . The electrical power dissipated (which determines brightness) is:

$P = \dfrac{V^2}{R}$

…

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CBSE 2020Set 55/1/11 markMCQ
$R_1$
›Reveal solutionSolution
In a parallel circuit, voltage is the same across both resistors, so power is inversely proportional to resistance. Since $P = V^2 / R$ , the ratio $P_1 : P_2 = R_2 : R_1 = 6 : 4 = 3 : 2$ . The correct option is (B).

The key to this problem is understanding what stays constant when resistors are in parallel. Many students jump to using $P = I^2 R$ without checking whether current is the same — that formula works only when the current through each resistor is identical, which is true in series but not in parallel.

In a parallel connection, the voltage across each resistor is the same (the battery voltage). That’s the anchor. So the natural formula to use is $P = \frac{V^2}{R}$ , because $V$ is common to both.

Let’s walk through it.

Identify the fixed quantity.

$R_1 = 4\ \Omega$ and $R_2 = 6\ \Omega$ are in parallel across the same battery. The voltage $V$ across each is identical.

Choose the right power formula.

Power dissipated in a resistor is $P = \frac{V^2}{R}$ . Since $V$ is the same for both, the power is inversely proportional to resistance:

$P_1 = \frac{V^2}{R_1}, \quad P_2 = \frac{V^2}{R_2}$

Write the ratio.

$P_1 : P_2 = \frac{V^2}{R_1} : \frac{V^2}{R_2} = \frac{1}{R_1} : \frac{1}{R_2} = R_2 : R_1$

Substitute the values. $P_1 : P_2 = 6 : 4 = 3 : 2$ …
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CBSE 2020Set 55/3/11 markMCQ
$\dfrac{V}{2}$
›Reveal solutionSolution
The power consumed by a resistor depends on the square of the applied voltage. Halving the voltage reduces the power to one-fourth of the original value, so the answer is $\dfrac{P}{4}$ .

Concept and Intuition

This problem tests a fundamental relationship in electricity: how power changes when voltage changes, assuming the resistance stays constant. The heater element is essentially a resistor — its resistance is fixed by its material and construction. When the manufacturer rates it as $(P, V)$ , they mean: "If you connect this heater to a $V$ volt supply, it will dissipate $P$ watts of power."

The key insight is that resistance doesn't change when you change the voltage. So we first find the resistance from the rated values, then use that resistance to compute the new power at the reduced voltage.

Watch out

A common mistake is to assume power is directly proportional to voltage. It's not — power depends on the square of voltage for a fixed resistor. Halving the voltage does NOT halve the power; it quarters it.

Step-by-Step Solution

Write the power formula for a resistor. For a resistor of resistance $R$ , the power dissipated when connected to a voltage $V$ is:

$P = \frac{V^2}{R}$
This comes from combining Ohm's law $V = IR$ with $P = VI$ .

Find the resistance from the rated values. The heater is rated $(P, V)$ , meaning at voltage $V$ it consumes power $P$ . So:

$P = \frac{V^2}{R} \quad \Rightarrow \quad R = \frac{V^2}{P}$
This resistance is a property of the heater element and does not change.

Now connect it to a source of voltage $\dfrac{V}{2}$ . …
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Q.Case Study : The following figure shows a circuit diagram. We can find the currents through and potential differences across the different resistors using Kirchhoff's rules. Answer the following questions based on the above :

Setting up the circuit

Part (a)

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