Ideal Source In an ideal source, all the internal energy or power, e.g., the chemical reaction in a battery, is available to the load without any “losses” in the form of heat. An ideal battery is shown symbolically as

An ideal voltage source (battery) in series with a load (light bulb): Find the following: –The current through the bulb whose resistance is R b = 2.5 Ω. –The power consumed by the light bulb in W. –The power supplied by the ideal battery in W.

The current drawn by the bulb is I = V source / R bulb = 1.5 / 2.5 = 0.6 A The power consumed by the bulb is P bulb = I 2 R bulb = (0.6 A) = 0.9 W The power supplied by the ideal battery is P source = V I = 1.5 (0.6) = 0.9 W From the law of the conservation of energy, P source = P bulb Solutions

Real Source Real sources have an internal resistance. This is not a visible resistor like you used in the lab, but it is a “model resistor” added to the circuit to analyze the source realistically. The internal resistance causes energy loss (dissipated heat for example) and causes the source to provide less energy to the load.

Ideal source (battery) in series with a load (light bulb) Real source (battery with internal resistance) in series with a load (light bulb)

Repeat the previous exercise with a real 1.5 V battery source. Assume the battery’s internal resistance to be 0.3Ω. Find the following: Current through the bulb Power consumed by the bulb Power supplied by the ideal portion of the battery Power consumed by the battery’s internal resistance.

Current: I = 1.5 V / 2.8 Ω = A Power consumed by the bulb: P bulb = I 2 R bulb = ( A) Ω = W Power provided by the ideal portion of the battery: P source = V I = 1.5V( A) = W Power consumed by the battery’s internal resistance: P internal = I 2 R internal = ( A) Ω = W R eq = R internal + R b = = 2.8 Ω

Comparison with Previous Results Ideal Source: I = 0.6 A P bulb = 0.9 W P source = 0.9 W Real Source: I = A P bulb = 0.717W P source = 0.803W P internal = W (dissipated as heat) Balance of power: P source = P internal + P bulb 0.803W = 0.086W W Balance of power: P source = P bulb 0.9 W = 0.9 W

Connection to Real Life As batteries age, their internal resistance increases. That means, even if you measure 1.5 V across the terminals with a meter, when you connect it to a load, it will provide a lower net voltage

In-Class Activity Given this circuit where the real battery has been connected for a while. Assume the bulb needs at least 0.4 A to light. What will the internal resistance be when the bulb goes dark? What voltage will be measured across the bulb at that time?

Can the bulb be replaced by a different bulb to maximize the power delivered to the bulb for a given battery voltage and internal resistance (R i ) ? In other words, if you replace the bulb with a symbolic resistance R b as shown here, can you find the value of R b for maximum bulb power? Start by expressing the power absorbed by the bulb in terms of V and both R’s. Question

Power Absorbed by Bulb P b = I b * V b I b = (in terms of R b, R i and V s ) V b = (in terms of R b, R i and V s ) Therefore, P b =

The power absorbed by the light bulb is given by where P b = bulb power (W) R b = bulb resistance (Ω) R i = internal resistance (Ω) V s = battery voltage (V) Is there a value of R b that maximizes P b ?

Use Excel to generate a table of P b as a function of R b for V = 1.5 V and R i = 0.3 Ω. Choose values of R b from 0 Ω to 0.8 Ω in 0.05 Ω steps. Plot your data using the “Scatter Plot” Graphical Approach In-Class Activity

Names: Team: Date: R_bulbP_bulb V s = 1.5 V R i = 0.3 Ω What is P b for R b = 0? What is P b for R b → ∞ ? From the curve, identify what value of R b makes P b maximum.

In calculus you’ve learned about derivatives. The first derivative of a function is a measure of its slope (rise/run). When we take the derivative of our power function with respect to R b, we find that the slope is zero at the maximum. The value of R b at that maximum will be R b = R i Analytical Approach