In the circuit shown in the figure (Figure 1) the batteries have negligible internal resistance and the meters are both idealized.

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In the circuit shown in the figure (Figure 1) the batteries have negligible internal resistance and the meters are both idealized.

In the circuit shown in the figure (Figure 1) the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 13.0 V.

Find the emf epsilon of the battery.

What will the ammeter read when the switch is closed?

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Answer 1

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Answer #1

with the switch open, the 13 V battery can be ignored.

That leaves us with 2 loops the left & the right hand one.

current through 50R = 13 /50 = 0.26A

voltage across 30 + 50 ohms = 80 * 026 = 20.8 V

This voltage also across 75R which has current through it of
i across 75 = 20.8/ 75 = 0.277A

total current through both loops = 0.26+ 0.277 = 0.537A

voltage drop across 20 R = 0.537* 20 = 10.74 V

total V = 20.8 + 10.74 = 31.54 V

---------------------------------------------

B.When the switch is closed, the voltage across the 50 ohm resister must be 25 volts
so the ammeter will read

I = V/R
I = 25 / 50
I = 0.5 Amps

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Answer 2

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In the circuit shown in the figure (Figure 1) the batteries have negligible internal resistance and the meters are both idealized.

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Add Answer to: In the circuit shown in the figure (Figure 1) the batteries have negligible internal resistance and the meters are both idealized.

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In the circuit shown in the figure (Figure 1) the batteries have negligible internal resistance and the meters are both idealized.