3.9. ELECTRICAL ENERGY POWER

Consider a conductor with end points A and B, in which a current I is flowing from A to B. The electric potential at A and B are denoted by V(A) and V(B) respectively. Since current is flowing from A to B, V(A) > V(B) and the potential difference across AB is V = V(A) – V(B) > 0.

This chunk explains the basic relationship of voltage and current in a conductor. When a current I flows through a conductor between two points A and B, point A has a higher electric potential than point B. The difference in electric potential (voltage) is expressed as V, which is the difference between the voltages at points A and B. This is crucial as it establishes the direction of the current flow.

Think of a water system where water flows from a higher elevation (point A) to a lower elevation (point B). Just as water flows downhill due to gravity, electric current flows from a high potential to a low potential.

In a time interval Dt, an amount of charge DQ = I Dt travels from A to B. The potential energy of the charge at A, by definition, was Q V(A) and similarly at B, it is Q V(B). Thus, change in its potential energy DU is DU = Final potential energy – Initial potential energy = DQ[(V(B) – V(A)] = –DQ V = –I V Dt < 0.

As charge moves from point A to point B, its potential energy decreases. The change in potential energy (DU) is calculated as the difference between its potential energy at A and at B. This change is negative, indicating that energy is lost from the charge, which is consistent with the flow of current moving from high potential to low potential.

Imagine a car moving down a hill. The car loses potential energy as it slides down the slope, just as the electric charge loses potential energy when moving from a point of higher voltage to a point of lower voltage.

Thus, in case charges were moving freely through the conductor under the action of electric field, their kinetic energy would increase as they move. We have, however, seen earlier that on the average, charge carriers do not move with acceleration but with a steady drift velocity. This is because of the collisions with ions and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms. The atoms vibrate more vigorously, i.e., the conductor heats up. Thus, in an actual conductor, an amount of energy dissipated as heat in the conductor during the time interval Dt is, DW = I V Dt. The energy dissipated per unit time is the power dissipated P = DW/Dt and we have, P = I V.

In reality, charges in a conductor do not move freely due to collisions with atoms, leading to energy dissipation as heat. The power dissipated, or the rate at which energy is used, is given by the product of the current (I) and the potential difference (V). This illustrates how electrical energy is converted into thermal energy as charges move through the conductor.

Think of a toaster. When electrical current passes through its wires, it encounters resistance, generating heat. This is the power (P) being converted from electrical energy to thermal energy. The heat is what cooks the bread inside the toaster.

Using Ohm’s law V = IR, we get P = I^2 R = V^2/R as the power loss (“ohmic loss”) in a conductor of resistance R carrying a current I.

This equation shows how power loss due to resistance in electrical circuits can be calculated. This loss, known as 'ohmic loss', is an important consideration in electrical systems, especially over long distances where transmission cables have resistance. It highlights the relationship between current, voltage, and resistance in determining power loss.

Consider the power lines that transport electricity over long distances. To reduce power loss, high voltages are used, which allows for lower currents and thus minimizes the I^2R losses. This is similar to how water pipes can leak if they are not pressurized enough.

Where does the power come from? As we have reasoned before, we need an external source to keep a steady current through the conductor. It is clearly this source which must supply this power. In the simple circuit shown with a cell, it is the chemical energy of the cell which supplies this power for as long as it can.

This section emphasizes the role of external energy sources in electrical circuits. For current to continue flowing, energy must be supplied, often by chemical reactions in batteries or electrical energy from power plants. This energy is converted into electrical energy, allowing charge carriers to move and perform work.

Think of a battery-powered toy. As long as the batteries have chemical energy to convert into electrical energy, the toy will function. When the battery runs out, the energy source is depleted, and the toy stops working.

We shall see now how this can be achieved. Consider a device R, to which a power P is to be delivered via transmission cables having a resistance R to be dissipated by it finally. If V is the voltage across R and I the current through it, then P = VI. The power dissipated in the connecting wires, which is wasted, is P_c with P_c = I^2 R_c, P_c = P^2 R_c/V^2.

This part discusses strategies for efficient power transmission. By increasing the voltage (V) used to transmit power, the current (I) can be reduced, which in turn minimizes the power loss in the lines due to resistance. This principle is why electrical power is transmitted at high voltages over long distances.

Think about how water is pumped through a city. To ensure it reaches the furthest faucets without losing pressure, pumps maintain pressure by using thicker pipes (analogous to higher voltage). This allows water to flow efficiently to each outlet with minimal loss.

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