Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 32, Problem 82CP

(a)

To determine

The current in the circuit The current in the circuit a long time after the switch has been in position a.

(a)

Expert Solution
Check Mark

Answer to Problem 82CP

The current in the circuit a long time after the switch has been in position a is 1.0A.

Explanation of Solution

Write the expression to obtain the current in the circuit The current in the circuit a long time after the switch has been in position a.

    I=εR1

Here, I is the current in the circuit, ε is the voltage across the battery and R1 is the resistance which is in series with the battery.

Conclusion:

Substitute 12.0V for ε and 12.0Ω for R1 in the above equation to calculate I.

    I=12.0V12.0Ω=1.0A

Therefore, the current in the circuit a long time after the switch has been in position a is 1.0A.

(b)

To determine

The initial voltage across each resistor and across the inductor when switch is thrown from position a to b.

(b)

Expert Solution
Check Mark

Answer to Problem 82CP

The initial voltage across the resistor which is in series with the battery is 12.0V and the initial voltage across the resistor which is in parallel with the battery is 1200V and the initial voltage across the inductor is 1212V when switch is thrown from position a to b.

Explanation of Solution

When switch is thrown from position a to b, the current in both the resistor and inductor remain same.

Write the expression to obtain the voltage across the resistor which is in series with the battery.

    VR1=IR1                                                                                                                (I)

Here, VR1 is the voltage across the resistor which is in series with battery, I is the current in  the circuit and R1 is the resistor which is in series with the battery.

Write the expression to obtain the voltage across the resistor which is in parallel with the battery.

    VR2=IR2                                                                                                               (II)

Here, VR2 is the voltage across the resistor which is in parallel with battery, I is the current in  the circuit and R2 is the resistor which is in parallel with the battery.

Write the expression to obtain the voltage across the inductor.

    VL=VR1+VR2                                                                                                         (III)

Here, VL is the voltage across the inductor, VR1 is the voltage across the resistor which is in series with battery and VR2 is the voltage across the resistor which is in parallel with battery.

Conclusion:

Substitute 1.0A for I and 12.0Ω for R1 in equation (I) to calculate VR1.

    VR1=(1.0A)(12.0Ω)=12.0V

Substitute 1.0A for I and 1200Ω for R2 in equation (II) to calculate VR2.

    VR2=(1.0A)(1200Ω)=1200V

Substitute 12.0V for VR1 and 1200V for VR2 in equation (III) to calculate VL.

    VL=12.0V+1200V=1212V

Therefore, the initial voltage across the resistor which is in series with the battery is 12.0V and the initial voltage across the resistor which is in parallel with the battery is 1200V and the initial voltage across the inductor is 1212V when switch is thrown from position a to b.

(c)

To determine

The time elapsed before the voltage across the inductor drops to 12.0V.

(c)

Expert Solution
Check Mark

Answer to Problem 82CP

The time elapsed before the voltage across the inductor drops to 12.0V is 7.6ms.

Explanation of Solution

Write the expression to obtain the time elapsed before the voltage across the inductor drops to 12.0V.

    ε=VL(e(R1+R2)tL)

Here, ε is the potential across the battery, VL is the potential across the inductor, R1 is the resistor which is in series with the battery, R2 is the resistor which is in parallel with the battery, L is the inductance of the inductor and t is the time elapsed.

Conclusion:

Substitute 12.0Ω for R1, 1200Ω for R2, 12.0V for ε, 2.00H for L and 1212V for VL in the above equation to calculate t.

    12.0V=1212V(e(12.0Ω+1200Ω)t2.00H)(e(12.0Ω+1200Ω)t2.00H)=12.0V1212V(e(12.0Ω+1200Ω)t2.00H)=0.0099((12.0Ω+1200Ω)t2.00H)=ln(0.0099)

Further solve the above equation.

    ((12.0Ω+1200Ω)t2.00H)=ln(0.0099)606t=4.61522t=4.61522606=0.0076s

Further solve the above equation.

    t=0.0076s×1000ms1s=7.6ms

Therefore, the time elapsed before the voltage across the inductor drops to 12.0V is 7.6ms.

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Chapter 32 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

Ch. 32 - Prob. 6OQCh. 32 - Prob. 7OQCh. 32 - Prob. 1CQCh. 32 - Prob. 2CQCh. 32 - Prob. 3CQCh. 32 - Prob. 4CQCh. 32 - Prob. 5CQCh. 32 - Prob. 6CQCh. 32 - The open switch in Figure CQ32.7 is thrown closed...Ch. 32 - Prob. 8CQCh. 32 - Prob. 9CQCh. 32 - Prob. 10CQCh. 32 - Prob. 1PCh. 32 - Prob. 2PCh. 32 - Prob. 3PCh. 32 - Prob. 4PCh. 32 - Prob. 5PCh. 32 - Prob. 6PCh. 32 - Prob. 7PCh. 32 - Prob. 8PCh. 32 - Prob. 9PCh. 32 - Prob. 10PCh. 32 - Prob. 11PCh. 32 - Prob. 12PCh. 32 - Prob. 13PCh. 32 - Prob. 14PCh. 32 - Prob. 15PCh. 32 - Prob. 16PCh. 32 - Prob. 17PCh. 32 - Prob. 18PCh. 32 - Prob. 19PCh. 32 - Prob. 20PCh. 32 - Prob. 21PCh. 32 - Prob. 22PCh. 32 - Prob. 23PCh. 32 - Prob. 24PCh. 32 - Prob. 25PCh. 32 - Prob. 26PCh. 32 - Prob. 27PCh. 32 - Prob. 28PCh. 32 - Prob. 29PCh. 32 - Prob. 30PCh. 32 - Prob. 31PCh. 32 - Prob. 32PCh. 32 - Prob. 33PCh. 32 - Prob. 34PCh. 32 - Prob. 35PCh. 32 - Prob. 36PCh. 32 - Prob. 37PCh. 32 - Prob. 38PCh. 32 - Prob. 39PCh. 32 - Prob. 40PCh. 32 - Prob. 41PCh. 32 - Prob. 42PCh. 32 - Prob. 43PCh. 32 - Prob. 44PCh. 32 - Prob. 45PCh. 32 - Prob. 46PCh. 32 - Prob. 47PCh. 32 - Prob. 48PCh. 32 - Prob. 49PCh. 32 - Prob. 50PCh. 32 - Prob. 51PCh. 32 - Prob. 52PCh. 32 - Prob. 53PCh. 32 - Prob. 54PCh. 32 - Prob. 55PCh. 32 - Prob. 56PCh. 32 - Prob. 57PCh. 32 - Prob. 58PCh. 32 - Electrical oscillations are initiated in a series...Ch. 32 - Prob. 60APCh. 32 - Prob. 61APCh. 32 - Prob. 62APCh. 32 - A capacitor in a series LC circuit has an initial...Ch. 32 - Prob. 64APCh. 32 - Prob. 65APCh. 32 - At the moment t = 0, a 24.0-V battery is connected...Ch. 32 - Prob. 67APCh. 32 - Prob. 68APCh. 32 - Prob. 69APCh. 32 - Prob. 70APCh. 32 - Prob. 71APCh. 32 - Prob. 72APCh. 32 - Prob. 73APCh. 32 - Prob. 74APCh. 32 - Prob. 75APCh. 32 - Prob. 76APCh. 32 - Prob. 77APCh. 32 - Prob. 78CPCh. 32 - Prob. 79CPCh. 32 - Prob. 80CPCh. 32 - Prob. 81CPCh. 32 - Prob. 82CPCh. 32 - Prob. 83CP
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