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Question: You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 Ω and C = 5.00 x 10-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

Short Answer

Expert verified

Answer:

(a) The emf on the battery is 44.0 V

(b) After 1.20×10-5sthe switch is closed the charge on the capacitor is equal to 40.0×10-6C

(c) When the current has magnitude then the energy being stored in the capacitor is 24W and supplied by the battery is 132 W

Step by step solution

01

EMF of the battery

Electromotive force (Emf) is the total amount of electrical potential energy a battery can supply before it is connected to an external circuit

02

Current through the circuit

Given,

C=5.00×10-6F, R=12.0Ωand time t=0 .

Firstly, we need to find the emf of the battery when the current has a magnitude of I= 3.00 A and the charge on the capacitor isQ=40.0×10-6C

Therefore, the charge over the capacitor is given by the equation:

Q=Cε(1-e-tτ)

Now, the current in the circuit is given by:

I=dQdt=εRe-tτ

03

Calculating the emf

Dividing the above two equations we get:

QA=CR1-e-tτe-tτ

Multiply byetτetτwe get:

QI=CRetτ-1

Solving for t we get:

t=τlnQICR+1

Substituting the values, we get,

t=(6.00×10-5s)ln40.0×10-6C(3.00A)(12.0Ω)(5.00×10-6F)+1=1.20×10-5s

Now, substitute the values in,

ε=IRetτ

We get,

ε=(3.00A)×(12.0Ω)×e1.20×10-5s6.00×10-5s=44.0V

Therefore, the emf is 44.0 V

04

Energy stored in both capacitors

From the above parts, we see that the current in the circuit has a magnitude of I =3.00 A and the charge on the capacitor is Q=40.0×10-6C, now the time at which the circuit has the above magnitude and charge is given by:

t=1.20×10-5s

Now the energy stored is given by:

role="math" localid="1655816376742" UC=Q22C

Therefore, the rate of energy stored in the capacitor is:

PC=QICPC=(40.0×10-6C)(3.00A)5.00×10-6F=24.0W

Therefore, the rate of energy stored in the capacitor is

Again, the rate of energy supplied by the battery is:

Pε=Iε=(3.00A)(44.0V)=132W

Therefore, the energy supplied by the battery is 132 W

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