Find initial current and resistance in a circuit

[moenste]

b(ii) gives what resistance ?

(b) (ii) E = I (r + R) → 12 = 4 (from the answer) * (0.5 + R) → R = 2.5 Ω.

In b(iii) you go off the rails, I'm afraid. The 10.5 V is over resistance X, NOT over the series circuit of X + 0.5 Ω !

So the resistor X has a voltage of 10.5 V?

 

[BvU]

The source gives off 12 V, after the internal resistor there is 10.5 V left over. What does that mean for the voltage drop over that internal resistor ? So what is the current through the series circiut of internal resistor and X ? THe voltage over X is given, so what is the resistance X ?

 

[moenste]

The source gives off 12 V, after the internal resistor there is 10.5 V left over. What does that mean for the voltage drop over that internal resistor ? So what is the current through the series circiut of internal resistor and X ? THe voltage over X is given, so what is the resistance X ?

We have a voltage battery 12 V. We have a resistor -2 V. We have a voltmeter 10 V. We have a resistor X 0.5 V and current I = 0.4 A. So R = V / I = 0.5 / 0.4 = 1.25 Ohm.

The source gives off 12 V, after the internal resistor there is 10.5 V left over. What does that mean for the voltage drop over that internal resistor ? So what is the current through the series circiut of internal resistor and X ? THe voltage over X is given, so what is the resistance X ?

12 - 10.5 = 0.5 I
I = 3 A new current in the parallel circuit?

 

[BvU]

We have a voltage battery 12 V. We have a resistor -2 V. We have a voltmeter 10 V. We have a resistor X 0.5 V and current I = 0.4 A. So R = V / I = 0.5 / 0.4 = 1.25 Ohm.

You confuse me. The X resistor sees 10 V. That's what the meter indicates. How can it possibly be 0.5 V !? Where does the 0.4 A come from ?

 

[moenste]

You confuse me. The X resistor sees 10 V. That's what the meter indicates. How can it possibly be 0.5 V !? Where does the 0.4 A come from ?

So the V_X = 10 V. And internal resistance of the battery is 0.5 Ohm. How do we find the R_X?

(i) 12 - 10 = 0.5 I → I = 4 A.

(ii) 12 = 4 (0.5 + R) → R = 2.5 Ω

(iii) 10.5 = 4 (0.5 + R) → R = 2.1 Ω

or 12 - 10.5 = 0.5 I → I = 3 A
10.5 = 3 (0.5 + R) → R = 3 Ω

or 12 - 10.5 = 0.5 I → I = 3 A
12 = 3 (0.5 + R) → R = 3.5 Ω

 

[cnh1995]

i) 12 - 10 = 0.5 I → I = 4 A.

(ii) 12 = 4 (0.5 + R) → R = 2.5 Ω

Right.

or 12 - 10.5 = 0.5 I → I = 3 A
12 = 3 (0.5 + R) → R = 3.5

Right.
Do you conceptually understand what is happening here? You got the temperature part right. If you can explain the drop in voltage from 12V to 10V, you have understood the concept of internal resistance of a voltage source.

 

[moenste]

Do you conceptually understand what is happening here? You got the temperature part right. If you can explain the drop in voltage from 12V to 10V, you have understood the concept of internal resistance of a voltage source.

The battery always has 12 V. Only the voltmeter changes from 10 V to 10.5 V.

And then we just use E = Ir + V and E = I (r + R), where E = 12 V, r = 0.5 Ohm and V is 10 V and 10.5 V.

 

[cnh1995]

The battery always has 12 V. Only the voltmeter changes from 10 V to 10.5 V.

Right. Internal resistance comes in series with the circuit and some voltage drop takes place across it.