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A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?

Short Answer

Expert verified

The chemical energy of the battery reduced isE=1.1×104J

Step by step solution

01

Given

Currenti=5ATimet=6minEmfε=6V

02

Determining the concept

Write an expression for the change in the energy using the definition of emf. Then, using the definition of current, replace the change in charge in the earlier expression and after inserting the given values, find the reduced chemical energy of the battery

Electromotive forceis defined as the electric potential produced by either electrochemical cell or by changing the magnetic field.

Formulae are as follow:

ε=dWdqi=qt

Where,

ε is emf, i is current, q is charge, w is work done, t is time.

03

Determining how much  the chemical energy of the battery is reduced

Emf of the battery is defined as work done per unit charge and is given as,

ε=WqW=εq

5A Current is up in the circuit for time t=6min. Then, the change in work done is nothing but the change in chemical energy W=E. Then,

E=εq

From the definition of current,

i=qtq=it

Hence,

E=t

Substituting all values in the above equation,

E=5A×6V×360sE=10800~1.1×104J

Hence, the chemical energy of the battery reduced isE=1.1×104J

Therefore, by using the definition of emf and the definition of current the change in energy can be determined.

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Most popular questions from this chapter

In Fig. 27-55a, resistor 3 is a variable resistor and the ideal battery has emf.ε=12V Figure 27-55b gives the current I through the battery as a function of R3. The horizontal scale is set by.R3s=20ΩThe curve has an asymptote of2.0 mAasR3. What are (a) resistanceR1and (b) resistance R2?

In Fig. 27-82, an ideal battery of emf ε=12.0Vis connected to a network of resistancesR1=12.0Ω, R2=12.0Ω,R3=4.0Ω,R4=3.00ΩandR5=5.00Ω. What is the potential difference across resistance 5?

In Fig. 27-58, a voltmeter of resistance RV=300Ωand an ammeter of resistanceRA=3.00Ω are being used to measure a resistance R in a circuit that also contains a resistance R0=100Ωand an ideal battery with an emf of ε=12.0V. ResistanceR is given by R=V/i, whereV is the potential across Rand iis the ammeter reading. The voltmeter reading is V', which is V plus the potential difference across the ammeter. Thus, the ratio of the two-meter readings is not R but only an apparent resistance role="math" localid="1664348614854" R'=V/i. If R=85.0Ω, what are (a) the ammeter reading, (b) the voltmeter reading, and (c)R' ? (d) IfRA is decreased, does the difference betweenR' andR increase, decrease, or remain the same?

(a) In Fig. 27-18a, are resistorsR1and R3in series?

(b) Are resistors R1&R3in parallel?

(c) Rank the equivalent resistances of the four circuits shown in Fig. 27-18, greatest first.

In Figure,ε1=3.00V,ε2=1.00V , R1=4.00Ω, R1=2.00Ω , R1=5.00Ω and both batteries are ideal. (a) What is the rate at which energy is dissipated in R1 ? (b) What is the rate at which energy is dissipated in R2? (c) What is the rate at which energy is dissipated in R3? (d) What is the power of battery 1? (e) What is the power of battery 2?

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