   Chapter 18, Problem 59AP

Chapter
Section
Textbook Problem

A voltage ΔV is applied to a series configuration of n resistors, each of resistance R. The circuit components are reconnected in a parallel configuration, and voltage ΔV is again applied. Show that the power consumed by the series configuration is 1/n2 times the power consumed by the parallel configuration.

To determine
The charges q1 and q2 on capacitors C1 and C2 after the switch is closed as a function of time.

Explanation

Given Info:

The first resistance R1 is 2.0 .

The second resistance R2 is 3.0

The first capacitance C1 is 2.0μF .

The second resistance C2 is 3.0μF

The total resistance in the circuit is 1.2 .

The total capacitance in the circuit is 5.0μF .

The total capacitance in the circuit is 5.0μF .

The emf of the battery is 120.0V .

Explanation:

Since, the resistances are in parallel connection,

Formula to calculate the total resistance is,

R=11R1+1R2

• R1 is the first resistance
• R2 is the second resistance

Substitute 2.0 for R1 and 3.0 for R2 to find the total resistance of the circuit,

R=112.0+13.0=1.2

Thus, the total resistance in the circuit is 1.2 .

Since, the capacitances are in parallel connection,

Formula to calculate the total capacitance is,

C=C1+C2

• C1 is the capacitance of the first capacitor
• C2 is the capacitance of the second capacitor

Substitute 2.0μF for C1 and 3.0μF for C2 to find the total capacitance of the circuit,

C=2.0μF+3.0μF=5.0μF

Thus, the total capacitance in the circuit is 5.0μF .

Formula to calculate the time constant of the circuit is,

τ=RC

• R is the total resistance of the circuit
• C is the total capacitance of the circuit

Substitute 1.2 for R and 5.0μF for C to find the time constant of the circuit,

τ=(1.2×103Ω)(5.0×106F)=6.0×103s=6.0s1000

Thus, the time constant of the circuit is 6.0s1000 .

Formula to calculate the maximum charge is,

Qmax=Cε

• ε is the emf of the battery
• C is the total capacitance of the circuit

Substitute 5

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