Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 28, Problem 28.74AP

An ideal voltmeter connected across a certain fresh 9-V battery reads 9.30 V, and an ideal ammeter briefly connected across the same battery reads 3.70 A. We say the battery has an open-circuit voltage of 9.30 V and a short-circuit current of 3.70 A. Model the battery as a source of emf ε in series with an internal resistance r as in Figure 27.1a. Determine both (a) ε and (b) r. An experimenter connects two of these identical batteries together as shown in Figure P27.45. Find (c) the open-circuit voltage and (d) the short-circuit current of the pair of connected batteries. (e) The experimenter connects a 12.0-Ω resistor between the exposed terminals of the connected batteries. Find the current in the resistor. (f) Find the power delivered to the resistor. (g) The experimenter connects a second identical resistor in parallel with the first. Find the power delivered to each resistor. (h) Because the same pair of batteries is connected across both resistors as was connected across the single resistor, why is the power in part (g) not the same as that in part (f)?

Figure P27.45

Chapter 28, Problem 28.74AP, An ideal voltmeter connected across a certain fresh 9-V battery reads 9.30 V, and an ideal ammeter

(a)

Expert Solution
Check Mark
To determine
The emf of the battery.

Answer to Problem 28.74AP

The emf of the battery is 9.30V .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A .

In an open circuit current the current of the battery is 0A .

Formula to calculate the emf of the battery is,

ε=V+IR

Here,

ε is the emf of the battery.

V is the voltage in the battery.

I is the current produced in the battery.

R is the resistance in the battery.

Substitute 9.30V  for V , 0A for I in the above equation.

ε=9.30V+(0A)(R)=9.30V

Conclusion:

Therefore, the emf of the battery is 9.30V .

(b)

Expert Solution
Check Mark
To determine
The resistance of the battery.

Answer to Problem 28.74AP

The resistance of the battery is 2.51Ω .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A .

Formula to calculate the internal resistance of the battery is,

ΔV=VIr=0

Here,

ΔV is the voltage difference of the battery.

I is the current produced in the battery.

r is the internal resistance in the battery.

Substitute 9.30V  for V , 3.70A for I in the above equation.

9.30V(3.70A)r=09.30V=(3.70A)rr=9.30V3.70A=2.51Ω

Conclusion:

Therefore, resistance of the battery is 2.51Ω .

(c)

Expert Solution
Check Mark
To determine
The open circuit voltage of the battery.

Answer to Problem 28.74AP

The open circuit voltage of the battery is 18.6V .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A . The batteries are connected in series. The voltages of the both batteries are same.

Formula to calculate the total emf of the battery is,

εT=2ε

Here,

εT is the total emf of the battery.

Substitute 9.30V  for ε in the above equation.

εT=2(9.30V)=18.6V

The total emf of the battery is equal to the open circuit voltage of the battery.

VT=18.6V

Conclusion:

Therefore, the open circuit voltage of the battery is 18.6V .

(d)

Expert Solution
Check Mark
To determine
The short circuit current of the pair of connected batteries.

Answer to Problem 28.74AP

The short circuit current of the pair of connected batteries is 3.70A .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A . The batteries are connected in series. The voltages of the both batteries are same.

The total resistance in the battery is,

rT=2r

Here,

rT is the total resistance in the battery.

Substitute 2.51Ω for r in the above equation.

rT=2(2.51Ω)=5.02Ω

Thus, the internal resistance of the battery is 5.02Ω .

Formula to calculate the short circuit current of the batteries is,

I=VTrT

Here,

VT is the open circuit voltage of the battery.

Substitute 18.6V  for VT , 5.02Ω for rT in the above equation.

I=18.6V5.02Ω=3.70A

Conclusion:

Therefore, the short circuit current of the pair of connected batteries is 3.70A .

(e)

Expert Solution
Check Mark
To determine
The current in the resistor if the 12.0Ω resistor is connected between the exposed terminals of the connected battery.

Answer to Problem 28.74AP

The current in the resistor 12.0Ω is 1.09A .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A . The batteries are connected in series. The voltages of the both batteries are same.

The total series resistance in the battery is,

Req=rT+12.0Ω

Here,

Req is the total series resistance in the battery.

Substitute 5.02Ω for rT in the above equation.

Req=5.02Ω+12.0Ω=17.02Ω

Thus, the total series resistance of the battery is 17.02Ω .

Formula to calculate the current in the resistor 12.0Ω is,

I1=VTReq

Here,

I1 is the current in the resistor 12.0Ω .

Substitute 18.6V  for VT , 17.02Ω for Req in the above equation.

I1=18.6V17.02Ω=1.092A1.09A

Conclusion:

Therefore, the current in the resistor 12.0Ω is 1.09A .

(f)

Expert Solution
Check Mark
To determine
The power delivered to the resistor.

Answer to Problem 28.74AP

The power delivered to the resistor is 14.3W .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A . The batteries are connected in series. The voltages of the both batteries are same.

Formula to calculate the power delivered to the resistor is,

P=(I1)2R

Here,

P is the power delivered to the resistor.

R is the resistance of the resistor.

Substitute 1.09A for I1 , 12.0Ω for R in the above equation.

P=(1.09A)212.0Ω=14.25W14.3W

Conclusion:

Therefore, the power delivered to the resistor is 14.3W .

(g)

Expert Solution
Check Mark
To determine
The power delivered to each resistor.

Answer to Problem 28.74AP

The power delivered to each resistor is 8.54W .

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A .

The batteries are connected in series. The voltages of the both batteries are same.

The equivalent internal resistance in the battery is,

R1=R'+rT=[(1R)+(1R)]+rT

Here,

R1 is the total resistance of the resistor.

R' is the equivalent internal resistance in parallel.

Substitute 12.0Ω for R , 5.02Ω for rT in the above equation.

R1=[(112.0Ω)+(112.0Ω)]+5.02Ω=6.00Ω+5.02Ω=11.02Ω11.0Ω

Thus, the total resistance of the resistor is 11.0Ω .

Formula to calculate the current in the batteries is,

I2=VTR1

Here,

I2 is the power delivered to the resistor.

Substitute 18.6V  for VT , 11.0Ω for R1 in the above equation.

I2=18.6V11.0Ω=1.69A

Thus, the current produced in the batteries is 1.69A .

Formula to calculate the terminal voltage across both batteries is,

ΔV'=εTI(rT)

Here,

ΔV' is the terminal voltage across both batteries.

Substitute 18.6V for εT , 1.69A for I2 , 5.02Ω for rT in the above equation.

ΔV'=18.6V1.69A(5.02Ω)=10.11V10.1V

Thus, the terminal voltage across both batteries is 10.1V .

Formula to calculate the power delivered to each resistor is,

P'=ΔV'R

Here,

P' is the power delivered to each resistor.

Substitute 10.1V for ΔV' , 12.0Ω for R in the above equation.

P'=(10.1V)212.0Ω=8.50W8.54W

Conclusion:

Therefore, the power delivered to each resistor is 8.54W .

(h)

Expert Solution
Check Mark
To determine
The reason for the power in part (g) is not same as in part (f).

Answer to Problem 28.74AP

The internal resistance of the batteries and the terminal voltage of the batteries is not same in both cases.

Explanation of Solution

Given info: The open circuit voltage of the battery is 9.30V . The short circuit current is 3.70A .

In part (g), the total internal resistance of the resistor is 11.0Ω and terminal voltage across both batteries is 10.1V . In part (f), the internal resistance is 12.0Ω . Thus, the power is dependent upon the current and resistance. So, if the resistance is change the power is also changed.

Conclusion:

Therefore, the internal resistance of the batteries and the terminal voltage of the batteries is not same in both cases.

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Chapter 28 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 28 - Prob. 28.6OQCh. 28 - What is the time constant of the circuit shown in...Ch. 28 - When resistors with different resistances are...Ch. 28 - When resistors with different resistances are...Ch. 28 - The terminals of a battery are connected across...Ch. 28 - Are the two headlights of a car wired (a) in...Ch. 28 - In the circuit shown in Figure OQ28.12, each...Ch. 28 - Prob. 28.13OQCh. 28 - A circuit consists of three identical lamps...Ch. 28 - A series circuit consists of three identical lamps...Ch. 28 - Suppose a parachutist lands on a high-voltage wire...Ch. 28 - A student claims that the second of two lightbulbs...Ch. 28 - Why is ii possible for a bird to sit on a...Ch. 28 - Given three lightbulbs and a battery, sketch as...Ch. 28 - Prob. 28.5CQCh. 28 - Referring to Figure CQ28.6, describe what happens...Ch. 28 - Prob. 28.7CQCh. 28 - (a) What advantage does 120-V operation offer over...Ch. 28 - Prob. 28.9CQCh. 28 - Prob. 28.10CQCh. 28 - A battery has an emf of 15.0 V. The terminal...Ch. 28 - Two 1.50-V batterieswith their positive terminals...Ch. 28 - An automobile battery has an emf of 12.6 V and 171...Ch. 28 - As in Example 27.2, consider a power supply with...Ch. 28 - Three 100- resistors are connected as shown in...Ch. 28 - Prob. 28.6PCh. 28 - What is the equivalent resistance of the...Ch. 28 - Consider the two circuits shown in Figure P27.5 in...Ch. 28 - Consider the circuit shown in Figure P28.9. Find...Ch. 28 - (a) You need a 45- resistor, but the stockroom has...Ch. 28 - A battery with = 6.00 V and no internal...Ch. 28 - A battery with emf and no internal resistance...Ch. 28 - (a) Kind the equivalent resistance between points...Ch. 28 - (a) When the switch S in the circuit of Figure...Ch. 28 - Prob. 28.15PCh. 28 - Four resistors are connected to a battery as shown...Ch. 28 - Consider die combination of resistors shown in...Ch. 28 - For the purpose of measuring the electric...Ch. 28 - Calculate the power delivered to each resistor in...Ch. 28 - Why is the following situation impossible? A...Ch. 28 - Consider the circuit shown in Figure P28.21 on...Ch. 28 - In Figure P28.22, show how to add just enough...Ch. 28 - The circuit shown in Figure P27.17 is connected...Ch. 28 - For the circuit shown in Figure P28.24, calculate...Ch. 28 - What are the expected readings of (a) the ideal...Ch. 28 - The following equations describe an electric...Ch. 28 - Taking R = 1.00 k and = 250 V in Figure P27.19,...Ch. 28 - You have a faculty position at a community college...Ch. 28 - The ammeter shown in Figure P28.29 reads 2.00 A....Ch. 28 - In the circuit of Figure P28.30, determine (a) the...Ch. 28 - Using Kirchhoffs rules, (a) find (he current in...Ch. 28 - In the circuit of Figure P27.20, the current I1 =...Ch. 28 - In Figure P28.33, find (a) the current in each...Ch. 28 - For the circuit shown in Figure P27.22, we wish to...Ch. 28 - Find the potential difference across each resistor...Ch. 28 - (a) Can the circuit shown in Figure P27.21 be...Ch. 28 - An uncharged capacitor and a resistor are...Ch. 28 - Consider a series RC circuit as in Figure P28.38...Ch. 28 - A 2.00-nF capacitor with an initial charge of 5.10...Ch. 28 - A 10.0-F capacitor is charged by a 10.0-V battery...Ch. 28 - In the circuit of Figure P27.25, the switch S has...Ch. 28 - In the circuit of Figure P27.25, the switch S has...Ch. 28 - The circuit in Figure P28.43 has been connected...Ch. 28 - Show that the integral 0e2t/RCdtin Example 27.11...Ch. 28 - A charged capacitor is connected to a resistor and...Ch. 28 - Prob. 28.46PCh. 28 - Prob. 28.47PCh. 28 - Turn on your desk lamp. Pick up the cord, with...Ch. 28 - Assume you have a battery of emf and three...Ch. 28 - Find the equivalent resistance between points a...Ch. 28 - Four 1.50-V AA batteries in series are used to...Ch. 28 - Four resistors are connected in parallel across a...Ch. 28 - The circuit in Figure P27.35 has been connected...Ch. 28 - The circuit in Figure P27.34a consists of three...Ch. 28 - For the circuit shown in Figure P28.55. the ideal...Ch. 28 - The resistance between terminals a and b in Figure...Ch. 28 - (a) Calculate the potential difference between...Ch. 28 - Why is the following situation impossible? A...Ch. 28 - A rechargeable battery has an emf of 13.2 V and an...Ch. 28 - Find (a) the equivalent resistance of the circuit...Ch. 28 - When two unknown resistors are connected in series...Ch. 28 - When two unknown resistors are connected in series...Ch. 28 - The- pair of capacitors in Figure P28.63 are fully...Ch. 28 - A power supply has an open-circuit voltage of 40.0...Ch. 28 - The circuit in Figure P27.41 contains two...Ch. 28 - Two resistors R1 and R2 are in parallel with each...Ch. 28 - Prob. 28.67APCh. 28 - A battery is used to charge a capacitor through a...Ch. 28 - A young man owns a canister vacuum cleaner marked...Ch. 28 - (a) Determine the equilibrium charge on the...Ch. 28 - Switch S shown in Figure P28.71 has been closed...Ch. 28 - Three identical 60.0-W, 120-V lightbulbs are...Ch. 28 - A regular tetrahedron is a pyramid with a...Ch. 28 - An ideal voltmeter connected across a certain...Ch. 28 - In Figure P27.47, suppose the switch has been...Ch. 28 - Figure P27.48 shows a circuit model for the...Ch. 28 - The student engineer of a campus radio station...Ch. 28 - The circuit shown in Figure P28.78 is set up in...Ch. 28 - An electric teakettle has a multiposition switch...Ch. 28 - A voltage V is applied to a series configuration...Ch. 28 - In places such as hospital operating rooms or...Ch. 28 - The switch in Figure P27.51a closes when Vc23Vand...Ch. 28 - The resistor R in Figure P28.83 receives 20.0 W of...
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