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Sunny_sXe [5.5K]
2 years ago
6

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula

te (a) the current in the circuit, (b) the terminal voltage of the battery, Vab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.
Physics
1 answer:
Luda [366]2 years ago
5 0

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

V_{ab} = i R

V_{ab} = (0.1832) (65)

V_{ab} = 11.91 Volts

(c)

Power dissipated in the resister R is given as

P_{R} = i²R

P_{R} = (0.1832)²(65)

P_{R} = 2.18 Watt

Power dissipated in the internal resistance is given as

P_{r} = i²r

P_{r} = (0.1832)²(0.5)

P_{r} = 0.0168 Watt

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Explanation:

From the Question we are told that

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Therefore

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Answer:

V_{a} - V_{b} = 89.3

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         V_{b} - V_{a} = - ∫ E ds

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We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

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            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

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As they ask us the reverse case

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             V_{a} - V_{b} = 89.3

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a.) 1 atm = 1.01325 bar

0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) =<em> 9.08×10⁻⁴ atm</em>

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Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at
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For example, when water boils at 100 ^{o}C then it changes into vapor state whereas when water freezes at 0^{0}C then it changes state from liquid to solid.

This means only physical state of water is changing and there is no change in chemical composition of water.

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faltersainse [42]

Answer:

83%

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