Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 28, Problem 70AP

(a)

To determine

The equilibrium charge on the capacitor as a function of R for the given circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 70AP

The equilibrium charge on the capacitor as a function of R for the given circuit is ((14.4×106F)((15×106R)F2.00Ω+R)).

Explanation of Solution

Let the resistance across 3.00Ω be R1, across 80.0Ω be R2 and across 2.00Ω be R3.

The resistors R1 and R2 are in series connection.

Write the expression for the equivalent resistance when the resistors are connected in series.

    Req=R1+R2                                                                                                              (I)

Here, the equivalent resistance is Req and the resistors connected in series are R1 and R2.

Write the expression to current through the series connection.

    I1=VReq                                                                                                                    (II)

Here, I1 is the current flowing through the resistors in series and V is the potential difference of the battery.

Write the expression to determine the potential difference across R2.

    V1=(I1R2)                                                                                                              (III)

Here, V1 is the potential difference across resistor R2.

The resistors R3 and R are in series connection.

Write the expression for the equivalent resistance when the resistors are connected in series.

    Req=R3+R                                                                                                           (IV)

Here, the equivalent resistance is Req.

Write the expression to current through the series connection.

    I2=VReq                                                                                                                   (V)

Here, I2 is the current flowing through the resistors in series.

Write the expression to determine the potential difference across R.

    V2=(I2R)                                                                                                               (VI)

Here, V2 is the potential difference across resistor R.

Write the expression to determine the potential difference across the capacitor.

    VC=|V1V2|                                                                                                          (VII)

Here, VC is the expression to determine the potential difference across the capacitor.

Write the expression to calculate the amount of charge stored in the capacitor.

    Q=VCC                                                                                                               (VIII)

Here, Q is the amount of charge stored in the capacitor and C is capacitance of the capacitor.

Conclusion:

Substitute 3.00Ω for R1 and 80.0Ω for R2 in equation (I) to solve for Req.

    Req=3.00Ω+80.0Ω=83.0Ω

Substitute 83.0Ω for Req and 5.00V for V in equation (II) to solve for I1.

    I1=5.0V83.0Ω=0.06A

Substitute 80.0Ω for R2, and 0.06A for I1 in equation (III) to solve for V1

    V1=((0.06A)80.0Ω)=4.80V

Substitute 2.00Ω for R3 in equation (IV) to solve for Req.

    Req=2.00Ω+R

Substitute 2.00Ω+R for Req and 5.00V for V in equation (V) to solve for I2.

    I2=5.00V2.00Ω+R

Substitute 5.00V2.00Ω+R for I2 in equation (VI) to solve for V2.

    V2=(5.00V2.00Ω+R(R))

Substitute (5.00V2.00Ω+R(R)) for V2 and 4.80V for V1 in equation (VII) to solve for VC.

    VC=|4.80V(5.00V2.00Ω+R(R))|

Substitute |4.80V(5.00V2.00Ω+R(R))| for VC and 3.00μF for C in equation (VIII) to solve for Q.

    Q=|4.80V(5.00V2.00Ω+R(R))|(3.00μF×106F1μF)=((14.4×106F)((15×106R)F2.00Ω+R))

Therefore, the equilibrium charge on the capacitor as a function of R for the given circuit is ((14.4×106F)((15×106R)F2.00Ω+R)).

(b)

To determine

The charge when R=10Ω.

(b)

Expert Solution
Check Mark

Answer to Problem 70AP

The charge when R=10Ω is 1.9μC.

Explanation of Solution

Write the expression for the equilibrium charge on the capacitor as a function of R for the given circuit.

    Q=((14.4×106F)((15×106R)F2.00Ω+R))                                                                (IX)

Conclusion:

Substitute 10.0Ω for R in equation (IX) to solve for Q.

    Q=((14.4×106F)(((15×106F)(10.0Ω))2.00Ω+10.0Ω))=1.9×106C×106μC1C=1.9μC

Therefore, the charge when R=10Ω is 1.9μC.

(c)

To determine

Whether the charge on the capacitor can be zero and the value of R when the charge is 0.

(c)

Expert Solution
Check Mark

Answer to Problem 70AP

Yes, the charge on the capacitor can be zero when the value of R is 48Ω.

Explanation of Solution

Write the expression for the equilibrium charge on the capacitor as a function of R for the given circuit.

    Q=((14.4×106F)((15×106R)F2.00Ω+R))                                                                 (X)

Conclusion:

Yes, the charge on the capacitor can be zero.

Substitute 0 for Q in equation (X) to solve for R.

    0=((14.4×106F)((15×106R)F2.00Ω+R))((15×106R)F2.00Ω+R)=(14.4×106F)(15×106R)F=(2.00Ω+R)(14.4×106F)(1514.4)×106R=28.8×106

Solve further.

    0.6R=28.8R=48Ω

Therefore, yes, the charge on the capacitor can be zero when the value of R is 48Ω.

(d)

To determine

The maximum possible value of the magnitude of charge and the value of R at which this is achieved.

(d)

Expert Solution
Check Mark

Answer to Problem 70AP

The maximum possible value of the magnitude of charge is 14.4μC and the value of R at which this is achieved is 0.00Ω.

Explanation of Solution

Write the expression for the equilibrium charge on the capacitor as a function of R for the given circuit.

    Q=((14.4×106F)((15×106R)F2.00Ω+R))                                                                 (XI)

It is clear from equation (XI) that the maximum charge can be achieved when the term containing R is reduced to zero.

This can be achieved by substituting zero for R.

Conclusion:

Substitute 0 for R in equation (XI) to solve for Qmax.

    Qmax=((14.4×106F)((15×106(0))F2.00Ω+(0)))=(14.4×106C×106μC1C)=(14.4μC)

Therefore, The maximum possible value of magnitude of charge is 14.4μC and the value of R at which this is achieved is 0.00Ω.

(e)

To determine

Whether it is experimentally meaningful to take R= and if yes, then the magnitude of charge if R=.

(e)

Expert Solution
Check Mark

Answer to Problem 70AP

It is experimentally not meaningful to take R=.

Explanation of Solution

Write the expression for the potential difference across R.

    V2=(5.00V2.00Ω+R(R))                                                                                           (XII)

Conclusion:

Substitute for R in equation (XII) to solve for V2.

    V2=(5.00V2.00Ω+())V2=

Thus, this infinite value of voltage across the resistor R shall lead to an infinite charge stored in the capacitor which is not possible.

Therefore, it is experimentally not meaningful to take R=.

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Chapter 28 Solutions

Physics for Scientists and Engineers With Modern Physics

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