3Ω 17 V A3 V3.

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JSIS E 111: Elementary Modern Greek

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JSIS E 111: Elementary Modern Greek

JSIS E 111: Elementary Modern Greek

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JSIS E 111: Elementary Modern Greek

JSIS E 111: Elementary Modern Greek

JSIS E 111: Elementary Modern Greek

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JSIS E 111: Elementary Modern Greek

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JSIS E 111: Elementary Modern Greek

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3Ω 17 V A3 V3.

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3Ω 17 V A3 V3.

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Μεταγράφημα παρουσίασης:

3Ω 17 V A3 V3

7Ω A3 V1 29Ω 31Ω 9Ω V3 A1 37 V A4 23Ω 7Ω V2 7Ω 8Ω A2 3Ω 5Ω

7Ω A3 V1 29Ω 31Ω 9Ω V3 A1 37 V A4 23Ω 7Ω V2 7Ω 8Ω A2 3Ω 5Ω

7Ω A3 V1 29Ω 31Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

7Ω A3 V1 29Ω 31Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

7Ω V1 14.9833Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

7Ω V1 14.9833Ω A1 37 V 23Ω 7Ω V2 19Ω 8Ω A2 5Ω

7Ω V1 21.9833Ω A1 37 V 23Ω 19Ω 8Ω A2 5Ω

7Ω V1 21.9833Ω A1 37 V 23Ω 19Ω 8Ω A2 5Ω

Finally we have this last series circuit: 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

Solving: Rtot =7+7.0622+5+8=27.0622Ω 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A (the reading on A1) 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A (the reading on A1) V1 = IR = 1.3672*7 = 9.5705V 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

which is the voltage across the subcircuit. Solving: Rtot =7+7.0622+5+8=27.0622Ω I = V/R = 37/27.0622 = 1.3672A (the reading on A1) V1 = IR = 1.3672*7 = 9.5705V And finally, V7.0622Ω = IR = 1.3672*7.0622 = 9.6556 V which is the voltage across the subcircuit. 7Ω V1 A1 37 V 7.0622Ω 8Ω 5Ω

This is what you have left: 14.9833Ω 9.6556 V 23Ω 7Ω V2 19Ω A2

This is what you have left: Solving for A2 is simple: I = V/R So A2 = 9.6556/23 = .4198A 14.9833Ω 9.6556 V 23Ω 7Ω V2 19Ω A2

Which leaves: 14.9833Ω 9.6556 V 7Ω V2 19Ω

Let’s look at the middle circuit 14.9833Ω 9.6556 V 7Ω V2 19Ω

This is what it really is 29Ω 31Ω 9.6556 V 7Ω V2

But let’s go back to: 14.9833Ω 9.6556 V 7Ω V2

We can solve this series circuit for V2 : But let’s go back to: We can solve this series circuit for V2 : Rtot = 7 + 14.9833 = 21.9833Ω 14.9833Ω 9.6556 V 7Ω V2

We can solve this series circuit for V2 : But let’s go back to: We can solve this series circuit for V2 : Rtot = 7 + 14.9833 = 21.9833Ω I = V/R = 9.6556/21.9833 = .4392A 14.9833Ω 9.6556 V 7Ω V2

We can solve this series circuit for V2 : But let’s go back to: We can solve this series circuit for V2 : Rtot = 7 + 14.9833 = 21.9833Ω I = V/R = 9.6556/21.9833 = .4392A V2 = IR = .4392*7 = 3.0746V And the voltage across the 14.9833Ω resistor is IR = .4392*14.9833 = 6.5810V 14.9833Ω 9.6556 V 7Ω V2

Now, just for fun, let’s look at the 14.9833Ω subcircuit 9.6556 V 7Ω V2

Now, just for fun, let’s look at the 14.9833Ω subcircuit Remember that it has a voltage of 6.5810V across it 14.9833Ω 9.6556 V 7Ω V2

So it looks like this: A3 29Ω 31Ω 6.5810V

We can find the current through the 31Ω resistor fairly simply: I = V/R = 6.5810/31 = .2123A 29Ω 31Ω 6.5810V

But this is not the only current running through the ammeter A3. We can find the current through the 31 resistor fairly simply: I = V/R = 6.5810/31 = .2123A But this is not the only current running through the ammeter A3. A3 29Ω 31Ω 6.5810V

It also reads the current going through the 19Ω subcircuit 29Ω 31Ω 9.6556 V .2123A 7Ω V2 19Ω

It also reads the current going through the 19Ω subcircuit I 19Ω = V/R = 9.6556/19 = .5082A A3 29Ω 31Ω 9.6556 V .2123A .5082A 7Ω V2 19Ω

It also reads the current going through the 19Ω subcircuit I 19Ω = V/R = 9.6556/19 = .5082A So the reading on A3 is the sum of these currents: = .5082+.2123 = .7205A A3 29Ω 31Ω 9.6556 V .2123A .5082A 7Ω V2 19Ω

Ta Daa! It also reads the current going through the 19Ω subcircuit I 19Ω = V/R = 9.6556/19 = .5082A So the reading on A3 is the sum of these currents: = .5082+.2123 = .7205A Ta Daa! A3 29Ω 31Ω 9.6556 V .2123A .5082A 7Ω V2 19Ω