Problem 21 - Technical Mathematics with Calculus

Question

From Eq. \(1066, P=I^{2} R,\) we see that power \(P\) dissipated in a resistor varies directly as the square of the current \(I\) in the resistor. By what factor must the current in an electric heating coil be increased to triple the power consumed by the heater?

Step-by-Step Solution

Answer

The current must be increased by a factor of sqrt{3} (approximately 1.732) to triple the power consumed by the heater.

Step 1 - Understand the relationship between power and current

Recognize that power ( P) is proportional to the square of the current ( I), given by the equation P = I^2 R, where R is the resistance of the heater which remains constant. To find the factor by which the current needs to be increased to triple the power, we will use this direct square relationship.

Step 2 - Set up the equation for triple power

Let the original power be P and the new power be 3 P (since we want to triple it). The corresponding currents are I (original current) and xI (new current, where x is the factor by which the current is increased). Using the power equation, we get the new power as (xI)^2 R = 3P. Since P = I^2 R, we can substitute P into the equation.

Step 3 - Solve for the current increase factor

With the equation (xI)^2 R = 3(I^2 R), simplify to find the factor x. Dividing both sides by I^2 R, we get x^2 = 3. Take the square root of both sides to solve for x, giving us x = sqrt{3}.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in the field of electronics and electrics; it defines the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. According to Ohm's Law,
\( V = I \times R \).
This equation states that the voltage across a resistor is directly proportional to the current flowing through it, with the resistance being the constant of proportionality.

• For a constant resistance, if you increase the voltage, the current will also increase proportionally.
• If the resistance is increased while keeping the voltage constant, the current will decrease.

In practical terms, it means that for a given resistor, if we know any two of these values, we can calculate the third. Our exercise about increasing the power consumed by a heater by adjusting the current is very much related to this law. We should keep the resistance of the heater constant to ensure that the relationship between the power and current follows the pattern given by the power formula and Ohm’s Law.

Resistance in Electrical Circuits

Resistance is the property of a material that opposes the flow of electric current. It is an essential factor in determining how much current will flow through a circuit for a given voltage. Resistance is measured in ohms (Ω).

• Conductors have low resistance and allow current to flow easily.
• Insulators have high resistance and impede the flow of current.
• Resistors are components specifically designed to provide a certain amount of resistance in a circuit.

In the context of our original exercise, we treat the resistance of the electric heating coil as a constant. This constant resistance is crucial because it ensures the predictability of the coil’s response to changes in current, in accordance with Ohm's Law. As the current increases, the power also rises, demonstrating their direct relationship based on the constant resistance of the heating element.

Electrical Power Formula

The formula for electrical power encapsulates how energy is converted in an electrical circuit. The power dissipated in a resistor, for example, can be calculated using the formula:
\[ P = I^2 R \],
where P stands for power in watts, I represents current in amperes, and R symbolizes the resistance in ohms. This formula shows that the power dissipated (or consumed) in a resistor is proportional to the square of the current through it, multiplied by the resistance.

Relating Power to Current

Following this formula, if you desire to adjust the power, you need to consider how changes in the current will affect it. To triple the power \( P \) in our original exercise, logically, the current must be adjusted in a way that the square of the current multiplied by the resistance becomes three times the original power—which involves finding the square root of 3. This illustrates the quadratic relationship between power and current when resistance is constant.