EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 25, Problem 103P

(a)

To determine

The battery current just after closing switch S.

(a)

Expert Solution
Check Mark

Answer to Problem 103P

The battery current just after closing the switch S is 0.250A .

Explanation of Solution

Given:

The value of emf is ε=50.0V .

The value of capacitances are C=5.00μF .

Formula used:

Apply Kirchhoff’s rule in circuit just after switch is closed,

  εI0(200Ω)VC=0 ..... (1)

Here, I0 is initial current in battery just after closing the switch.

Calculation:

Initially, the capacitor is uncharged so,

  VC=0

From equation (1), the battery current just after closing switch S is calculated as,

  (50.0V)I0(200Ω)0=0I0(200Ω)=50.0VI0=0.250A

Conclusion:

Therefore, the battery current just after closing the switch S is 0.250A .

(b)

To determine

The battery current a long time after closing the switch S.

(b)

Expert Solution
Check Mark

Answer to Problem 103P

The battery current a long time after closing the switch S is 62.5mA .

Explanation of Solution

Formula used:

Apply Kirchhoff’s rule in circuit a long time after switch is closed,

  εI(200Ω)I(600Ω)=0 ..... (2)

Here, I is initial current in battery a long time after closing the switch.

Calculation:

From equation (2), the battery current a long time after closing switch S is calculated as,

  (50.0V)I(200Ω)I(600Ω)=0I(800Ω)=50.0VI=62.5×103AI=62.5mA

Conclusion:

Therefore, the battery current a long time after closing the switch S is 62.5mA .

(c)

To determine

The current in 600Ω resistor as a function of time.

(c)

Expert Solution
Check Mark

Answer to Problem 103P

The current in 600Ω resistor as a function of time is (62.5mA)(1et/0.500ms) .

Explanation of Solution

Formula used:

Apply Kirchhoff’s rule at j unction of resistor 200Ω and capacitors 5.00μF ,

  I1=I2+I3 ..... (3)

Apply Kirchhoff’s rule in loop 1,

  εQCR1I1=0 ..... (4)

Apply Kirchhoff’s rule in loop containing resistor 600Ω and capacitors 5.00μF ,

  QCR2I2=0 ..... (5)

Calculation:

Differentiate equation (4) with respect to time,

  ddt(εQCR1I1)=001CdQdtR1dI1dt=01CI3R1dI1dt=0R1dI1dt=I3C ..... (6)

Differentiate equation (5) with respect to time,

  ddt(QCR2I2)=01CdQdtR2dI2dt=0R2dI2dt=I3C ..... (7)

From equation (3) and (7),

  R2dI2dt=( I 1 I 2 )CdI2dt=( I 1 I 2 )R2C ..... (8)

From equation (4),

  I1=εQ/CR1=εR2I2R1 ..... (9)

From equation (8) and (9),

  dI2dt=1R2C( ε R 2 I 2 R 1 I2)=εR1R2CR1+R2R1R2CI2

Let the solution of above differential equation is,

  I2=a+bet/τ ..... (10)

Differentiate equation (10) with respect to time,

  dI2dt=bτet/τ ..... (11)

From equation (8) and (11),

  bτet/τ=εR1R2C(R1+R2R1R2C)(a+bet/τ)

Equate coefficient of et/τ of above equation on both sides,

  bτ=( R 1 + R 2 R 1 R 2 C)bτ=R1R2CR1+R2 ..... (12)

And,

  a=εR1+R2

At t=0 , I2=0 ,

  a+b=0b=ab=εR1+R2

From equation (10),

  I2=εR1+R2εR1+R2et/τ ..... (13)

From equation (10),

  τ=( 200Ω)( 600Ω)( 5.00μF)( 200Ω)( 600Ω)=5×106s=0.500ms

Substitute values in equation (13),

  I2=50.0V200Ω+600Ω(1e t/ 0.500ms )=(62.5mA)(1e t/ 0.500ms )

Conclusion:

Therefore, the current in 600Ω resistor as a function of time is (62.5mA)(1et/0.500ms) .

(d)

To determine

The charges on capacitors plates a long time after reopening the switch S.

(d)

Expert Solution
Check Mark

Answer to Problem 103P

The charges on capacitors plates a long time after reopening the switch S is zero.

Explanation of Solution

Calculation:

If the switch S is reopened, then after long time there will not be any flow of current in the circuit. Thus,

  I=0

The potential difference across 10.0μF and 5.00μF capacitors will be zero,

  V10.0μF=V5.00μF=0

The charges on 10.0μF and 5.00μF capacitors will be zero,

  Q10.0μF=Q5.00μF=0

Conclusion:

Therefore, the charges on capacitors plates a long time after reopening the switch S is zero.

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Chapter 25 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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