Q.Using Kirchhoff's rules, calculate the current through the $40\,\Omega$ and $20\,\Omega$ resistors in the following circuit:

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

1. Find the resistance of each heater from its rating.

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

For heater 2 rated at $(P_2, V)$:

2. Calculate the total resistance in series.

   When connected in series, resistances add:

3. Apply the actual supply voltage.

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

Substituting:

4. Simplify the expression. …

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

1. 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:

Since $P_2 > P_1$, it follows that $R_2 < R_1$ (higher power means lower resistance).

2. Find the total resistance when they are in series.

Clearly $R_{\text{series}} > R_1$ (and also $> R_2$).

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

4. Compare $P_{\text{series}}$ with $P_1$. Since $P_1 > 0$, we have $P_1 + P_2 > P_2$, so $P_{\text{series}} = \frac{P_1 P_2}{P_1 + P_2} < \frac{P_1 P_2}{P_2} = P_1.$ …

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:

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. …

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.

1. 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.

2. 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:

3. Write the ratio.

4. Substitute the values. $P_1 : P_2 = 6 : 4 = 3 : 2$ …

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.

Step-by-Step Solution

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

This comes from combining Ohm's law $V = IR$ with $P = VI$.

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

This resistance is a property of the heater element and does not change.

3. Now connect it to a source of voltage $\dfrac{V}{2}$. …