Current through ammeter with two batteries

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Homework Statement


For the values shown in the figure below, calculate (a) the current through the ammeter A1, (b) the current through the ammeter A2, (c) the PD across R2.

242b0846387a.jpg


Assume that the internal resistance of each cell and the resistance of each ammeter is negligible.

Answers: (a) 0.05 A, (b) 0.0125 A, (c) 0.5 V

2. The attempt at a solution
(c) V = IR = 0.05 * 10 = 0.5 V

(a-b) We use Kirchhoff's rules: current flowing from E1 is I1, current entering R1 is I1 - I3 and current entering E2 is I3.

So we get 2 = 40 (I1 - I3) + 10 (1 - I3)
1.5 = - 40 (I1 - I3), minus since current is moving from left to right while we are moving in an anti-clockwise direction.

2 = 50 1 - 50 I3
1.5 = -40 I1 + 40 I3 (multiply by 1.25 and sum up)
3.875 = 0 which is uncorrect.

What am I doing wrong?
 

Answers and Replies

  • #2
cnh1995
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By observing the circuit, what can you say about the voltage across the 40 ohm resistor? What will be the voltage across the 10 ohm resistor?
 
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  • #3
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By observing the circuit, what can you say about the voltage across the 40 ohm resistor? What will be the voltage across the 10 ohm resistor?
Don't we need to find the current first? The voltage across the 10 Ω resistor is easily calculated when we know the current: (c) V = IR = 0.05 * 10 = 0.5 V, where I = 0.05 A is taken from the answers.
 
  • #4
gneill
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Don't we need to find the current first? The voltage across the 10 Ω resistor is easily calculated when we know the current
What component is connected directly across that resistor? What do you know about parallel-connected components?
 
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  • #5
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What component is connected directly across that resistor? What do you know about parallel-connected components?
Current is connected directly across the 10 Ohm resistor.

Parallel-connected components have the same voltage.
 
  • #6
cnh1995
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Parallel-connected components have the same voltage.
Right.
Current **Voltage source** is connected directly across the 10 Ohm resistor.
So what is the voltage across the 10 ohm resistor?
 
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  • #7
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Right.

So what is the voltage across the 10 ohm resistor?
Are you implying that we can just 2 V - 1.5 V = 0.5 V? But it is in (c), so I guess we need to find current in (a-b) first.
 
  • #8
cnh1995
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Are you implying that we can just 2 V - 1.5 V = 0.5 V? But it is in (c), so I guess we need to find current in (a-b) first.
Sorry I misread the post. Actually, voltage source E2 is directly connected across the 40 ohm resistor.
 
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  • #9
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Sorry I misread the post. Actually, voltage source E2 is directly connected across the 40 ohm resistor.
Since we have two batteries therefore we need to use Kirchhoff's rules, right?
 
  • #10
gneill
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Kirchhoff's rules always apply, so it's a pretty good bet they'll come in handy here!
 
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  • #11
cnh1995
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Since we have two batteries therefore we need to use Kirchhoff's rules, right?
The voltage source E2 is directly connected across the 40 ohm resistor. And you yourself said correctly that
Parallel-connected components have the same voltage.
What is the voltage across the 40 ohm resistor then?
 
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  • #12
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The voltage source E2 is directly connected across the 40 ohm resistor. And you yourself said correctly that

What is the voltage across the 40 ohm resistor then?
Ah, now I got it what you imply. Since we have a battery E2 = 1.5 V, then V40 Ω = E2 = 1.5 V. We can then find (c) 2 - 1.5 = 0.5 V.

But how do we find the currents? I = V / R = 2 / (40 + 10) = 0.04 A, not 0.05 A.
 
  • #13
cnh1995
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Ah, now I got it what you imply. Since we have a battery E2 = 1.5 V, then V40 Ω = E2 = 1.5 V. We can then find (c) 2 - 1.5 = 0.5 V.
Right.
2 / (40 + 10) = 0.04 A, not 0.05 A.
Ohm's law gives I=V/R where I is the current through R and V is the voltage across R.

 
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  • #14
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Right.

Ohm's law gives I=V/R where I is the current through R and V is the voltage across R.
We can use I = V / R for the R2 resistor. We'll get 0.5 / 10 = 0.05 A.

Then we find 1.5 / 40 = 0.0375 A. 0.05 - 0.0375 = 0.0125 A.
 
  • #15
cnh1995
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We can use I = V / R for the R2 resistor. We'll get 0.5 / 10 = 0.05 A.

Then we find 1.5 / 40 = 0.0375 A. 0.05 - 0.0375 = 0.0125 A.
Right.
 
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