   Chapter 18, Problem 61AP

Chapter
Section
Textbook Problem

A battery with an internal resistance of 10.0 Ω produces an open circuit voltage of 12.0 V. A variable load resistance with a range from 0 to 30.0 Ω is connected across the battery. (Note: A battery has a resistance that depends on the condition of its chemicals and that increases as the battery ages. This internal resistance can be represented in a simple circuit diagram as a resistor in series with the battery.) (a) Graph the power dissipated in the load resistor as a function of the load resistance. (b) With your graph, demonstrate the following important theorem: The power delivered to a load is a maximum if the load resistance equals the internal resistance of the source.

(a)

To determine
The equivalent resistance and current in each resistor when the power is connected between points A and B.

Explanation

Given Info:

The power supply is 5.0V .

The resistance R1 is 0.100Ω .

The resistance R2 is 1.0Ω .

The resistance R3 is 10.0Ω .

The total resistance of R2 , R3 and 100Ω in series connection is 111.0Ω .

The power supply is 5.0V .

Explanation:

The equivalent resistance is,

When the power supply is connected between A and B, the resistors R2 , R3 and 100Ω are in series. The total resistance of R2 , R3 and 100Ω are in parallel with the R1 resistance.

Thus, the equivalent resistance of the circuit is,

1Requ=1R1+1R2+R3+100Ω

• R1 is the first resistance
• R2 is the second resistance
• R3 is the third resistance

Substitute 0.100Ω for R1 , 1.0Ω for R2 and 10.0Ω for R3 to find the equivalent resistance.

Requ=110.100Ω+11.0Ω+10.0Ω+100Ω=110.100Ω+1111Ω=0.099Ω

The total resistance of R2 , R3 and 100Ω is 111

(b)

To determine
The equivalent resistance and current in each resistor when the power is connected between points A and C.

(c)

To determine
The equivalent resistance and current in each resistor when the power is connected between points A and D.

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