Current through ammeter with two batteries

[moenste]

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.

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?

 

[cnh1995]

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?

 

[moenste]

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.

 

[gneill]

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?

 

[moenste]

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.

 

[cnh1995]

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?

 

[moenste]

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.

 

[cnh1995]

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.

 

[moenste]

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?

 

[gneill]

Kirchhoff's rules always apply, so it's a pretty good bet they'll come in handy here!

 

[cnh1995]

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?

 

[moenste]

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 V_{40 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.

 

[cnh1995]

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.

 

[moenste]

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.

 

[cnh1995]

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.