Question

Show that the power supplied to the circuit in the figure by the battery with internal resistance is maximum when the resistance of the resistor in the circuit, \(R\), is equal to \(R_{i}\). Determine the power supplied to \(R\). For practice, calculate the power dissipated by a \(12.0-\mathrm{V}\) battery with an internal resistance of \(2.00 \Omega\) when \(R=1.00 \Omega, R=2.00 \Omega,\) and \(R=3.00 \Omega\)

Short answer

Answer: The maximum power is supplied to the resistor when the resistor's resistance is equal to the internal resistance of the battery (R = Ri).

Step-by-step solution

Calculate the overall current in the circuit

First, we need to calculate the overall current in the circuit, which can be done using Ohm's Law, \(I=V/(R+R_{i})\). Where \(I\) is the current, \(V\) is the voltage of the battery, \(R\) is the resistance of the resistor, and \(R_{i}\) is the internal resistance of the battery.

Determine the power to the resistor R

Now we will calculate the power supplied to resistor R using the formula \(P_{R}=I^{2}R\). We derived this formula from \(P=IV\), substituting \(I\) from Step 1.

Determine the maximum power to resistor R

To find the maximum power supplied to the resistor, we need to differentiate \(P_{R}\) with respect to \(R\) and set the result equal to zero. This will give us a condition for the value of \(R\) when the power is maximized.

Compute example power dissipation

Finally, let's verify our findings by plugging in the given values for a 12 V battery with an internal resistance of 2 Ω and different external resistances. Let's proceed with the calculations:

Calculate the overall current in the circuit

We'll use Ohm's Law for the general case: \(I=\frac{V}{(R+R_{i})}\)

Calculate the power to resistor R

Using the formula \(P_{R}=I^{2}R\), with the current found in the general case in step 1, we obtain \(P_{R}=(\frac{V}{(R+R_{i})})^{2}R\)

Determine the maximum power to resistor R

Differentiate \(P_{R}\) with respect to R and set it equal to zero: \(\frac{dP_{R}}{dR} = 0\) After solving this differential equation - using the quotient rule- and setting it to zero: \(R=R_{i}\). That means the maximum power is delivered when \(R=R_{i}\)

Compute example power dissipation

For this part of the problem, we will use the values of \(V=12.0V\) and \(R_{i}=2.0 \Omega\). Let's calculate the power dissipation for each value of external resistance \(R\): 1. \(R=1.0 \Omega\) \(P_{R1} = (\frac{12}{(1+2)})^{2} \cdot 1 = 2.67 W\) 2. \(R=2.0 \Omega\) \(P_{R2} = (\frac{12}{(2+2)})^{2} \cdot 2 = 3.6 W\) 3. \(R=3.0 \Omega\) \(P_{R3} = (\frac{12}{(3+2)})^{2} \cdot 3 = 2.88 W\) As we can see, the power supplied to \(R\) reaches its maximum when \(R = R_{i} = 2.0 \Omega\). This is consistent with our proof earlier in the solution, which demonstrated that maximum power is supplied when \(R = R_{i}\).

Key concepts

Ohm's Law

Ohm's Law is fundamental to understanding electrical circuits, setting the relationship between voltage (V), current (I), and resistance (R). This principle states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them, expressed by the formula:

\[I = \frac{V}{R}\]
Understanding Ohm's Law is crucial in circuit analysis as it allows us to determine one of these quantities if the other two are known, enabling us to calculate how current moves through a circuit which is essential for exercises like the one in our textbook problem. When working with problems involving internal resistance, it's vital to consider the total resistance in the circuit, that is the sum of external and internal resistance, to accurately apply Ohm's Law.

Circuit Analysis

Circuit analysis involves understanding how electrical components within a circuit interact to influence current flow and potential drops. In the context of the maximum power transfer theorem problem, analysis starts by identifying the total resistance in the circuit. This total resistance impacts the overall current, as seen in our textbook exercise, calculated through the combined resistances of the internal components and the load applied, or external resistance. By evaluating the effects of varying the external load, circuit analysis can reveal the conditions for maximum power dissipation in the load, integral to optimizing performance in electrical circuits and electronic devices.

Power Dissipation

Power dissipation refers to the process where electrical energy is converted into heat as the current flows through resistance. It's a significant concept when analyzing circuits as it ultimately determines the efficiency and safety of electrical systems. In resistors, power dissipation is governed by the formula:

\[P = I^2R\]
This shows that power dissipation is proportional to the square of the current multiplied by the resistance through which the current flows. The textbook exercise demonstrates how changing the resistance affects the power dissipated across it. The maximum power transfer theorem highlights the importance of power dissipation as it states that this value is maximized when the load resistance equals the source's internal resistance, a condition that we use to find the ideal setup for a circuit.

Internal Resistance

Internal resistance is an inherent resistance present within power sources, like batteries and generators, which affects the performance of a circuit. It acts as an obstacle to the flow of current in addition to any external resistance. The significance of internal resistance becomes evident especially when discussing the efficiency and behavior of a circuit under different load conditions. It contributes to the overall resistance used to calculate current via Ohm's Law, as seen in our textbook's example. When considering internal resistance, we recognize that a real source of voltage cannot provide unlimited power; it loses some energy due to its internal resistance, a concept that plays a crucial role in optimizing circuits for maximum power transfer.