Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 18, Problem 94P

(a)

To determine

Draw the equivalent circuit with one resistor and one capacitor.

(a)

Expert Solution
Check Mark

Answer to Problem 94P

The equivalent circuit with one resistor and one capacitor is shown below.

Explanation of Solution

The resistor R1 and R2 in series then the equivalent resistance is,

Req=R1+R2 (I)

The capacitors C1 and C2 are in parallel then the equivalent capacitor is,

C'=C1+C2

This capacitor is in series with C3. Then the final equivalent capacitance is,

Ceq=[1C'+1C3]1

Substitute the value of C' in the above equation to get the final equivalent capacitance.

Ceq=[1C1+C2+1C3]1 (II)

Conclusion:

Substitute 25 Ω for R1 and 33 Ω for R1 in equation I.

Req=25 Ω+33 Ω=58 Ω

Substitute 12 μF for C1 , 23 μF for C2 and 46 μF for C3 in equation II.

Ceq=[112 μF+23 μF+146 μF]1=20 μF

The equivalent resistance and capacitance values are used to draw the equivalent circuit and shown below.

Physics, Chapter 18, Problem 94P

(b)

To determine

Find the charge on the capacitor C1 and current on the resistor R1.

(b)

Expert Solution
Check Mark

Answer to Problem 94P

The charge on the capacitor C1 is 4.1×105 C and current on the resistor R1 is 0

Explanation of Solution

The current in the resistor with fully charged capacitor is zero. Since there is no flow of charge from the battery to the capacitor.

Write the equation for charge on equivalent capacitor,

Q=Ceq(ΔV) (III)

Here, Q is the charge, Ceq equivalent capacitance and ΔV is the potential difference.

Conclusion:

Substitute 6.0 V for ΔV and 20.0×106 F for Ceq in equation III.

Q=(20.0×106 F)(6.0 V)=1.2×104 C

This charge is same on the charge of capacitor C3 . Then find the voltage drop across C3.

V3=Q3C3

Substitute 1.2×104 C for Q3 and 46.0×106 F for C3 in the above equation.

V3=1.2×104 C46.0×106 F=2.61V

Then the voltage drop across the capacitors C1 and C2 is,

6.0 V2.61 V=3.39V

Then the charge on the capacitor C1 is,

Q1=C1V1

Substitute 3.39V for V1 and  12×106 F for C1 in the above equation.

Q1=(12×106 F)(3.39V)=4.1×105 C

Therefore, charge on the capacitor C1 is 4.1×105 C and current on the resistor R1 is 0.

(c)

To determine

Find the time constant of the circuit.

(c)

Expert Solution
Check Mark

Answer to Problem 94P

The time constant of the circuit is 1.2 ms.

Explanation of Solution

Write the equation for time constant,

τ=RC (IV)

Here, τ is the time constant, R is the resistance and C is the current.

Conclusion:

Substitute 58 Ω for R  and 20.0×106 F for C in equation IV.

τ=(58 Ω)(20.0×106 F)=1.2 ms

Therefore, the time constant of the circuit is 1.2 ms.

(d)

To determine

Find the time at which the voltage across the three capacitors 50% of its final value.

(d)

Expert Solution
Check Mark

Answer to Problem 94P

The time at which the voltage across the three capacitors 50% of its final value is 8.4×104 s.

Explanation of Solution

Write the equation for voltage across the capacitor.

V=ε(1et/(RC)) (V)

Here V is the voltage across the capacitor, ε is the emf or potential difference, t is the time, R is the resistance and C is the capacitance.

The voltage across the capacitor is equal to 50% of the emf.

Then, the voltage is V=0.5ε , substitute this in equation V.

0.5ε=ε(1et/(RC))0.5=1et/(RC)

Rewrite the above equation to get time t .

et/(RC)=0.50tRC=ln0.50

t=RCln0.50 (VI)

Conclusion:

Substitute 58 Ω for R  and 20.0×106 F for C in equation VI.

t=(58 Ω)(20.0×106 F)ln0.50=8.0×104 s

Therefore, the time at which the voltage across the three capacitors 50% of its final value is 8.4×104 s.

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Chapter 18 Solutions

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