All Exams Test series for 1 year @ ₹349 only
Question

A network consists of a large number of ideal linear resistors, one of which is R and 2 constant ideal sources. The power consumed by R is P1 when only the first source is active and P2 when only the second source is active. If both sources are active simultaneously, then the power consumed by R is

The correct answer is \({\left( {\sqrt {{P_1}} \pm \sqrt {{P_2}} } \right)^2}\)

Understanding Power Consumption in Resistors

In electrical circuits, the power consumed by a resistor is a fundamental concept. For an ideal linear resistor with resistance denoted as R, the power (P) dissipated can be calculated using its current (I) or voltage (V) with the following formulas:

  • P = I2R
  • P = V2/R

The question presents a scenario involving a resistor R within a network that includes two constant ideal sources. We are provided with the power consumed by R when each source operates independently: P1 when only the first source is active, and P2 when only the second source is active. The objective is to determine the total power consumed by R when both sources are simultaneously active.

Applying the Superposition Principle to Circuits

The superposition principle is highly effective for analyzing linear circuits. It states that in a linear circuit with multiple independent sources, the voltage across or the current through any element is the algebraic sum of the voltages or currents produced by each independent source acting individually, while all other sources are turned off (replaced by their internal impedances).

Let's focus on the current flowing through the resistor R:

  • When the first source is the only one active, let the current through R be I1. The power consumed is P1 = I12R.
  • When the second source is the only one active, let the current through R be I2. The power consumed is P2 = I22R.

From these power equations, we can establish relationships involving the currents:

  • Taking the square root, we get $\sqrt{{P_1}} = \sqrt{{{I_1}^2 R}} = |I_1|\sqrt{R}$. This means $|I_1| = \sqrt{P_1/R}$.
  • Similarly, $\sqrt{{P_2}} = \sqrt{{{I_2}^2 R}} = |I_2|\sqrt{R}$. This means $|I_2| = \sqrt{P_2/R}$.

When both sources are active simultaneously, the total current, Itotal, flowing through resistor R is the superposition of the individual currents. This means Itotal = I1 ± I2, where the sign (plus or minus) depends on whether the currents I1 and I2 flow in the same or opposite directions through R.

Calculating the Combined Power Consumption

The total power, Ptotal, consumed by resistor R when both sources are active is given by the formula Ptotal = Itotal2R.

Substituting the expression for Itotal:

$P_{total} = ({I_1} \pm {I_2})^2 R$

To express this result in terms of P1 and P2, we use the relationships derived earlier: $I_1 = \pm \sqrt{P_1/R}$ and $I_2 = \pm \sqrt{P_2/R}$. The specific signs depend on the actual polarities determined by the sources.

Thus, the total current can be represented as:

$I_{total} = (\pm \sqrt{P_1/R}) \pm (\pm \sqrt{P_2/R})$

This simplifies to:

$I_{total} = \frac{1}{\sqrt{R}} (\pm \sqrt{P_1} \pm \sqrt{P_2})$

Now, we calculate the total power using this expression for Itotal:

$P_{total} = (I_{total})^2 R = \left( \frac{1}{\sqrt{R}} (\pm \sqrt{P_1} \pm \sqrt{P_2}) \right)^2 R$

Expanding this equation:

$P_{total} = \frac{1}{R} (\pm \sqrt{P_1} \pm \sqrt{P_2})^2 R$

Which simplifies to:

$P_{total} = (\pm \sqrt{P_1} \pm \sqrt{P_2})^2$

The term $(\pm \sqrt{P_1} \pm \sqrt{P_2})$ encompasses the various combinations possible due to the relative directions of the currents contributed by each source.

  • The case $(\sqrt{P_1} + \sqrt{P_2})^2$ corresponds to the situation where the contributions to the current through R from both sources are in the same direction (additive).
  • The case $(\sqrt{P_1} - \sqrt{P_2})^2$ corresponds to the situation where the contributions are in opposite directions (subtractive).

Therefore, the power consumed by resistor R when both sources are active simultaneously is the square of the sum or difference of the square roots of the powers consumed when each source was active individually.

Final Answer Explanation

Based on the application of the superposition principle to the currents and the fundamental power formula $P = I^2 R$, the total power consumed by the resistor R when both sources are active concurrently is determined to be:

$P_{total} = {\left( {\sqrt {{P_1}} \pm \sqrt {{P_2}} } \right)^2}$

This result matches the third option provided in the question.

Was this answer helpful?

Important Questions from Network Theorems

  1. Which of the following is essential for reciprocity theorem to be applicable?
  2. A linear element satisfies the property (ies) of: 

  3. KVL gives the law of conservation of

  4. The maximum power transfer theorem is used in

  5. Millman’s theorem yields:

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App