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

A source maintains a current I in a resistor of resistance R. If V is the potential difference across the resistor, the electrical energy dissipated in the resistor in time t is given by _____.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

VIt

Calculating Electrical Energy Dissipation in a Resistor

Electrical energy is dissipated when current flows through a resistor. This process is often referred to as Joule heating or ohmic heating. The amount of energy dissipated depends on the voltage across the resistor, the current flowing through it, and the time duration for which the current flows.

Understanding Power and Energy Dissipation

Power (P) is the rate at which energy (E) is transferred or dissipated. Electrical power dissipated in a resistor is related to the potential difference (V) across it and the current (I) flowing through it by the formula:

\(P = VI\)

According to Ohm's Law, \(V = IR\). We can substitute this into the power formula to get other forms:

  • Substituting \(V = IR\) into \(P = VI\):
    \(P = (IR)I = I^2R\)
  • Substituting \(I = V/R\) into \(P = VI\):
    \(P = V(V/R) = \frac{V^2}{R}\)

So, the power dissipated in a resistor can be expressed in three equivalent ways:

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

Electrical energy (E) dissipated over a time (t) is the product of power and time:

\(E = P \times t\)

Deriving Energy Dissipation Formulas

Using the different expressions for power, we can find the energy dissipated in time t:

  • Using \(P = VI\):
    \(E = (VI)t = VIt\)
  • Using \(P = I^2R\):
    \(E = (I^2R)t = I^2Rt\)
  • Using \(P = \frac{V^2}{R}\):
    \(E = \left(\frac{V^2}{R}\right)t = \frac{V^2}{R}t\)

Therefore, the electrical energy dissipated in a resistor in time t can be given by \(VIt\), \(I^2Rt\), or \(\frac{V^2}{R}t\).

Analyzing the Options for Electrical Energy

Let's examine the given options in the context of the derived formulas for electrical energy dissipated in a resistor:

  • Option 1: \(V^2\) - This represents the square of the potential difference, not energy. It is not a valid formula for energy dissipation.
  • Option 2: \(IR^3t\) - This does not match any of the standard formulas (\(VIt\), \(I^2Rt\), \(\frac{V^2}{R}t\)). Let's check if it can be derived using Ohm's Law. If we substitute \(I = V/R\) into \(I^2Rt\), we get \((V/R)^2Rt = (V^2/R^2)Rt = V^2t/R\). If we substitute \(V = IR\) into \(VIt\), we get \((IR)It = I^2Rt\). This option \(IR^3t\) does not simplify to any of the standard energy formulas using Ohm's Law.
  • Option 3: \(VIt\) - This is one of the standard formulas for electrical energy dissipated, derived directly from \(E = Pt\) and \(P = VI\).
  • Option 4: \(VI^2t\) - This does not match any of the standard formulas. It seems like a mix-up of \(VIt\) and \(I^2Rt\).

Based on the standard formulas for electrical energy dissipation in a resistor, the expression \(VIt\) is a correct representation.

Conclusion on Electrical Energy Dissipation

The electrical energy dissipated in the resistor in time t, given the potential difference V across the resistor and the current I through it, is indeed \(VIt\). This aligns with the fundamental definition of power as \(VI\) and energy as power multiplied by time.

Revision Table: Electrical Energy and Power

QuantitySymbolStandard UnitFormulas for Resistor
Potential DifferenceVVolt (V)\(V = IR\)
CurrentIAmpere (A)\(I = V/R\)
ResistanceROhm (\(\Omega\))\(R = V/I\)
Power DissipatedPWatt (W)\(P = VI = I^2R = \frac{V^2}{R}\)
Energy DissipatedEJoule (J)\(E = Pt = VIt = I^2Rt = \frac{V^2}{R}t\)
TimetSecond (s)

Additional Information on Electrical Energy and Power

Understanding the relationship between voltage, current, resistance, power, and energy is crucial in electrical circuits. Here are some related concepts:

  • Ohm's Law: This fundamental law states that the potential difference (V) across a resistor is directly proportional to the current (I) flowing through it, provided the temperature and other physical conditions remain constant. Mathematically, \(V \propto I\), which leads to \(V = IR\), where R is the constant of proportionality, the resistance.
  • Joule's Law of Heating: This law describes the rate at which heat is produced in an electrical circuit. It states that the power (P) dissipated as heat in a resistor is directly proportional to the square of the current (I) flowing through it and the resistance (R). Mathematically, \(P = I^2R\). This is also known as Joule-Lenz Law. The total heat energy produced is then \(Q = P \times t = I^2Rt\). The formulas \(VIt\) and \(\frac{V^2}{R}t\) are equivalent expressions for this heat energy, derived using Ohm's Law.
  • Units: In the International System of Units (SI), potential difference is measured in Volts (V), current in Amperes (A), resistance in Ohms (\(\Omega\)), time in seconds (s), power in Watts (W), and energy in Joules (J). 1 Watt is equivalent to 1 Joule per second (1 W = 1 J/s).

These concepts help explain how electrical energy is converted into thermal energy when current flows through a resistive component, which is the basis for many electrical devices like heaters, incandescent light bulbs, and fuses.

Was this answer helpful?

Similar Questions

  1. Electric current is considered to be the flow of _________.

  2. When a number of resistors are connected in series in a circuit, the value of current ________ across each resistor.

  3. The resistance of a conductor is inversely proportional to:

  4. If a body takes ‘t’ seconds to go once around the circular path of radius ‘r’, the velocity ‘v’ is given by

  5. A current I flows through a resistor. A source maintains a potential difference of V across the resistor. The energy supplied by the source in time t is:

  6. If the resistance of a conductor is doubled, the current gets halved. This is because:

  7. In a Class 2 lever, effort and load move in the:

  8. Two identical resistors, each of 10 Ω, are connected in parallel. This combination, in turn, is connected to a third resistor in series of 10 Ω. The equivalent resistance of the combination is ________.

  9. If the power of a corrective lens in +2.0D, then it is a:

  10. Insulators have resistivity of the order of ________.


Important Questions from Physics

  1. What special name is given to the frictional force exerted by a fluid?

  2. ______ is used in periscope.

  3. Zero degree centigrade is equal to what degree Fahrenheit?

    A. 100°F

    B. 30°F

    C. 34°F

    D. 32°F   

  4. Keeping voltage constant, if more lamps are put into a series circuit, the overall current in the circuit:

    A. Increases

    B. Decreases

    C. Remains the same

    D. Becomes infinite

  5. Excessive curvature of eye lens leads to _______

Need Expert Advice?
Upcoming Exams
RRB NTPC
September 27, 2026
Test Series
RRB ALP img
Railways
RRB ALP 2026 Mock Test series
1035 Tests 1 Tests Free
886 Attempts
4.3(235)
English, Hindi
More Questions from RRB ALP

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