College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 18, Problem 61AP

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)

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

Answer to Problem 61AP

The equivalent resistance of the circuit is 0.099Ω .

The current through the resistor R1 is 50.0A and the current through the resistors R2 , R3 and 100Ω is 0.0450A .

Explanation of Solution

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,

College Physics:, Chapter 18, Problem 61AP , additional homework tip  1

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 111.0Ω .

Thus, the equivalent resistance of the circuit is 0.099Ω .

Formula to calculate the current is,

IR=VR

  • V is the voltage across the terminal
  • R is the equivalent resistance of the circuit

For resistor R1 ,

Substitute 5.0V for V and 0.100Ω for R1 to find the current through the resistor R1 .

IR1=5.0V0.100Ω=50.0A

Since, the resistors R2 , R3 and 100Ω are in a series connection; the current in these resistors will be the same.

For R2 , R3 and 100Ω ,

Substitute 5.0V for V and 111.0Ω for R to find the current through the resistors R2 , R3 and 100Ω ,

IR2=IR3=IR100=5.0V111.0Ω=0.0450A

Thus, the current through the resistor R1 is 50.0A and the current through the resistors R2 , R3 and 100Ω is 0.0450A .

Conclusion:

The equivalent resistance of the circuit is 0.099Ω .

The current through the resistor R1 is 50.0A and the current through the resistors R2 , R3 and 100Ω is 0.0450A .

(b)

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

Answer to Problem 61AP

The equivalent resistance of the circuit is 1.09Ω .

The current through the resistor R1 and R2 is 4.55A and the current through the resistors R3 and 100Ω is 0.0455A .

Explanation of Solution

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 , R1 is 1.10Ω .

The total resistance of R3 , R100 is 110.0Ω .

The power supply is 5.0V .

Explanation:

The equivalent resistance is,

College Physics:, Chapter 18, Problem 61AP , additional homework tip  2

When the power supply is connected between A and C, the resistors R1 , R2 are in series connection and the resistors R3 and. 100.0Ω are in series connection. The total resistance of R2 , R1 is in parallel with the total resistance of R1 and 100Ω resistance.

Thus, the equivalent resistance of the circuit is,

1Requ=1R1+R2+1R3+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+1.0Ω+110.0Ω+100Ω=111.10Ω+1110Ω=1.09Ω

The total resistance of R2 , R1 is 1.10Ω .

The total resistance of R3 , R100 is 110.0Ω .

Thus, the equivalent resistance of the circuit is 1.09Ω .

Formula to calculate the current is,

IR=VR

  • V is the voltage across the terminal
  • R is the equivalent resistance of the circuit

For resistor R1 and R2 ,

Substitute 5.0V for V and 1.10Ω for R to find the current through the resistors R1 and R2 ,

IR1=IR2=5.0V1.10Ω=4.55A

Since, the resistors R3 and 100Ω are in a series connection; the current in these resistors will be the same.

For R3 and 100Ω ,

Substitute 5.0V for V and 110.0Ω for R to find the current through the resistors R3 and 100Ω ,

IR3=IR100=5.0V110.0Ω=0.0455A

Thus, the current through the resistor R1 and R2 is 4.55A and the current through the resistors R3 and 100Ω is 0.0455A .

Conclusion:

The equivalent resistance of the circuit is 1.09Ω .

The current through the resistor R1 and R2 is 4.55A and the current through the resistors R3 and 100Ω is 0.0455A .

(c)

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

Answer to Problem 61AP

The equivalent resistance of the circuit is 9.99Ω .

The current through the resistor R3 , R1 and R2 is 0.450A and the current through the resistor 100Ω is 0.050A .

Explanation of Solution

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 , R1 and R3 is 11.1Ω .

The power supply is 5.0V .

Explanation:

The equivalent resistance is,

College Physics:, Chapter 18, Problem 61AP , additional homework tip  3

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

Thus, the equivalent resistance of the circuit is,

1Requ=1R1+R2+R3+1100Ω

  • 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+1.0Ω+10.0Ω+1100Ω=1111.1Ω+1100Ω=9.99Ω

The total resistance of R2 , R1 and R3 is 11.1Ω .

Thus, the equivalent resistance of the circuit is 9.99Ω .

Formula to calculate the current is,

IR=VR

  • V is the voltage across the terminal
  • R is the equivalent resistance of the circuit

For resistor R3 , R1 and R2 ,

Substitute 5.0V for V and 11.1Ω for R to find the current through the resistors R3 , R1 and R2 ,

IR1=IR2=IR3=5.0V11.1Ω=0.450A

For 100Ω ,

Substitute 5.0V for V and 100.0Ω for R to find the current through the resistor 100Ω ,

IR100=5.0V100.0Ω=0.050A

Thus, the current through the resistor R3 , R1 and R2 is 0.450A and the current through the resistor 100Ω is 0.050A .

Conclusion:

The equivalent resistance of the circuit is 9.99Ω .

The current through the resistor R3 , R1 and R2 is 0.450A and the current through the resistor 100Ω is 0.050A .

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