PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
10th Edition
ISBN: 9781337888462
Author: SERWAY
Publisher: CENGAGE L
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Chapter 27, Problem 45AP

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 27, Problem 45AP, 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 45AP

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 45AP

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 45AP

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 45AP

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 45AP

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 45AP

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 45AP

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 45AP

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 27 Solutions

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN

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