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Chapter 27, Problem 1P

Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into a flashlight. One battery has an internal resistance of 0.255 Ω, and the other has an internal resistance of 0.153 Ω. When the switch is closed, the bulb carries a current of 600 mA. (a) What is the bulb’s resistance? (b) What fraction of the chemical energy transformed appears as internal energy in the batteries?

(a)

Expert Solution
Check Mark
To determine

The value of the bulb resistance.

Answer to Problem 1P

The value of the bulb resistance is 4.59Ω .

Explanation of Solution

Given information: Internal resistance of one battery is 0.255Ω , internal resistance of the other battery is 0.153Ω , current when switch is closed is 600mA , voltage across the batteries is 1.50V .

`

Write the expression for the total emf for the circuit.

E=nV (1)

Here,

E is the total emf for the circuit.

n is the number of batteries.

V is the voltage across the batteries.

Substitute 2 for n , 1.50V for V in equation (1) to find E ,

E=2×1.50V=3.0V

Thus, the total emf for the circuit is 3.0V .

Formula to calculate the total internal resistance for the circuit.

r=r1+r2 (2)

Here,

r is the total internal resistance for the circuit.

r1 is the internal resistance of one battery

r2 is the internal resistance of other battery

Substitute 0.255Ω for r1 , 0.153Ω for r2 in equation (2) to find r ,

r=0.255Ω+0.153Ω=0.408Ω

Thus, the total internal resistance for the circuit is 0.408Ω .

Formula to calculate the value of the bulb resistance.

E=I(R+r)R=EIr (3)

Here,

I is the current when switch is closed.

R is the resistance of the bulb.

Substitute 3.0V for E , 0.408Ω for r , 600mA for I in equation (3) to find R ,

R=3.0V600mA×1A1000mA0.408Ω=4.59Ω

Thus, the value of the bulb resistance is 4.59Ω .

Conclusion:

Therefore, the value of the bulb resistance is 4.59Ω .

(b)

Expert Solution
Check Mark
To determine

The fraction of the chemical energy transformed appears as internal energy in the batteries.

Answer to Problem 1P

The fraction of the chemical energy transformed appears as internal energy in the batteries is 8.16% .

Explanation of Solution

Given information: Internal resistance of one battery is 0.255Ω , internal resistance of the other battery is 0.153Ω , current when switch is closed is 600mA , voltage across the batteries is 1.50V .

Formula to calculate the internal energy of the battery.

E=Ir (4)

Here,

E is the internal energy of the battery.

Substitute 0.408Ω for r , 600mA for I in equation (4) to find E ,

E=(600mA×1A1000mA)×0.408Ω=0.2448V0.245V

Thus, the internal energy of the battery is 0.245V .

Formula to calculate the fraction of the chemical energy transformed appears as internal energy in the batteries.

Ef=EE×100

Here,

Ef is the fraction of the chemical energy transformed appears as internal energy in the batteries.

Substitute 3.0V for E , 0.245V for E in equation (4) to find Ef ,

Ef=0.245V3.0V×100=8.16%

Thus the fraction of the chemical energy transformed appears as internal energy in the batteries is 8.16% .

Conclusion:

Therefore, the fraction of the chemical energy transformed appears as internal energy in the batteries is 8.16% .

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Chapter 27 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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