For the circuit shown in the figure, both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.45 A.

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For the circuit shown in the figure, both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.45 A.

For the circuit shown in the figure, both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.45 A.

a) What does the voltmeter read?
b) What is the emf ε of the battery?

science physics

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Answer 1

Answer #1

if the effective resistance across A and B is R, then

1/R = 1/25 + 1/15 + 1/(15+10) = 11/75

R = 75/11 ohms

Since, the ammeter reads 1.45 A, the potential difference between A and B is V= current x resistance = 1.45 x 25 = 36.25 volts

total current in the circuit I = V/R = 36.25 / (75/11) = 319/ 60 A

since this is the total current in the circuit, the same current will pass through the 45 ohm resistor.

So, potential difference across the 45 volts resistor = current x resistance = (319/60) x 45 = 957/4 volts

= 239.25 volts this will be the reading of the voltmeter.

b.) E = current x total resistance

E = (319/60) x ( 35 + 45 + R ) = 5.316666667 x ( 35 + 45 + 75/11 ) = 461.583333 volts

answered by: Tinsit

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Answer 2

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For the circuit shown in the figure, both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.45 A.

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For the circuit shown in (Figure 1) both meters are idealized, the battery has no appreciable internal resistance, and the ammeter reads 1.40 A.