Two equal resistances when connected in series to a battery consume electric power of 60 W. If these

Q: Why do we use the formula $P = \frac{V^2}{R}$ instead of $P = I^2R$?

In this problem, the resistors are connected to the *same battery* in both cases, meaning the voltage $V$ remains constant. The total current $I$, however, changes because the equivalent resistance changes. Using $P = \frac{V^2}{R}$ allows us to work with the constant $V$, making the math much simpler.

Q: What happens to the equivalent resistance when we switch from series to parallel?

In series, resistances add up, so the equivalent resistance increases ($R + R = 2R$). In parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals, which causes the overall resistance to decrease ($\frac{R}{2}$). A lower resistance allows more current to flow from the same voltage source.

Q: Why does the parallel circuit consume more power?

Power is inversely proportional to resistance when voltage is constant ($P = \frac{V^2}{R}$). Since the parallel combination has a much lower equivalent resistance ($\frac{R}{2}$) compared to the series combination ($2R$), it draws significantly more current from the battery, leading to a higher total power dissipation.

Q: Can I just add the individual powers in the parallel circuit?

Yes, absolutely! In a parallel circuit, each resistor gets the full battery voltage $V$. The power consumed by a single resistor is $\frac{V^2}{R}$. From our series calculation, we found $\frac{V^2}{R} = 120 \text{ W}$. Since there are two resistors in parallel, the total power is $120 \text{ W} + 120 \text{ W} = 240 \text{ W}$.

Q: Is there a direct formula relating series and parallel power for identical resistors?

Yes! For $n$ identical resistors connected to the same voltage source, the power consumed in parallel is $n^2$ times the power consumed in series. So, $P_p = n^2 P_s$. Here, $n=2$, so $P_p = 2^2 \times 60 = 240 \text{ W}$.

Question

Two equal resistances when connected in series to a battery consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be

Options:

(A) 60 W

(B) 30 W

(C) 240 W

(D) 120 W

Sigma Percentile · Accepted Answer

### The Power of Connections

Welcome to a classic problem that beautifully illustrates how the arrangement of components in an electrical circuit drastically alters its behavior. We are given two identical resistors. When they are connected in series across a battery, they consume $60 	ext{ W}$ of power. The question asks us to find the power they would consume if they were rewired in parallel across the exact same battery.

Let's break this down step-by-step and uncover the underlying physics.

### Analyzing the Series Circuit

Imagine the two identical resistors, each having a resistance $R$. When connected in series, the current has only one path to flow through. The equivalent resistance of this series combination is simply the sum of the individual resistances:

$$R_{eq} = R + R = 2R$$

The power $P$ consumed by a circuit can be expressed in multiple ways: $I^2R$, $V^2/R$, or $VI$. Since the battery voltage $V$ remains constant in both the series and parallel scenarios, the most strategic formula to use is $P = \frac{V^2}{R_{eq}}$.

For our series circuit, the power consumed is given as $60 	ext{ W}$:

$$P_s = \frac{V^2}{2R} = 60 	ext{ W}$$

From this equation, we can isolate a very useful term. By multiplying both sides by 2, we find the power that a *single* resistor would consume if connected directly to the battery:

$$\frac{V^2}{R} = 120 	ext{ W}$$

Keep this value locked in your memory; it is the key to unlocking the second half of the problem.

### The Parallel Shift

Now, let's rewire the circuit. We take the same two resistors and connect them in parallel across the same battery. In a parallel circuit, the current splits, and the equivalent resistance drops significantly. The new equivalent resistance $R'_{eq}$ is:

$$R'_{eq} = \frac{R 	imes R}{R + R} = \frac{R^2}{2R} = \frac{R}{2}$$

Notice how the equivalent resistance is now half of a single resistor's value, whereas in series, it was double!

### The Grand Finale

Let's calculate the new power consumed by this parallel combination. Using our trusty power formula again:

$$P_p = \frac{V^2}{R'_{eq}} = \frac{V^2}{R/2}$$

The $2$ in the denominator flips up to the numerator, giving us:

$$P_p = 2 \left(\frac{V^2}{R}ight)$$

Do you recognize the term in the parentheses? It's the exact value we isolated earlier! Substituting $\frac{V^2}{R} = 120 	ext{ W}$ into our equation yields:

$$P_p = 2 	imes 120 = 240 	ext{ W}$$

And there we have it! The power consumed in the parallel combination is $240 	ext{ W}$.

### A Pro-Tip for the Future

There is a brilliant shortcut you can use for competitive exams like JEE. If you have $n$ identical resistors connected across the same voltage source, the ratio of power consumed in parallel to the power consumed in series is always $n^2$.

$$P_{parallel} = n^2 	imes P_{series}$$

In our problem, $n = 2$. Therefore, $P_p = 2^2 	imes 60 = 4 	imes 60 = 240 	ext{ W}$. This trick saves precious time and acts as a fantastic cross-check for your calculations!

Animated Solution for Physics - Current Electricity: Two equal resistances when connected in series to a battery consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be

Select Answer:

Visualized Solution

$R_{e q}$ in Series

$Power in Series$

$\frac{V ^{2}}{R} = 120 W$

$R_{e q}^{'}$ in Parallel

$Power in Parallel$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Power of Connections

Analyzing the Series Circuit

The Parallel Shift

The Grand Finale

A Pro-Tip for the Future

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