# Battery charging circuit problem

1. Nov 20, 2012

### trollcast

1. The problem statement, all variables and given/known data
Circuit as shown below.

At what voltage of E does the battery begin to charge?

What percentage of the power is delivered to the 1R resistor by the voltage supply E when the charging current is zero?

What is the current through the battery when E = 20V?

2. Relevant equations

3. The attempt at a solution

I'm really not sure how to start with this one as there's nothing like this in the class notes as its an extension exercise the teacher set.

I assume for the first part E obviously must be greater than 12V so I need to work out the effect of the resistors in the circuit?

#### Attached Files:

• ###### Charging circuit.png
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2. Nov 20, 2012

### Staff: Mentor

If you place the reference (ground) node at the confluence of the resistors at the bottom, can you write the node equation for the top node?

What would you expect the current to be in the battery branch when it's just on the threshold of charging?

3. Nov 21, 2012

### trollcast

If the voltage at the top node is V then would it be:

$$\frac{V-12}{0.2} + \frac{V}{1} + \frac {V - E}{0.4} = 0$$

I'm not sure about this as its not something I've learnt before but I'd assume it would be very small just above 0?

4. Nov 21, 2012

### Staff: Mentor

Yes, good. Solve for V in terms of E (or E in terms of V).

A tip: Sometimes it's esthetically pleasing to work towards a solution comprised of whole numbers or fractions and without decimals. In this case you might replace 0.2 with 1/5 and 0.4 with 2/5 for a neat derivation.
Yup. A charging battery has current flowing INTO its positive terminal. The threshold, then, occurs when the current is zero and any increase in applied potential would cause current to start to flow.

What value must V have in order for the battery branch current to be zero?

5. Nov 21, 2012

### trollcast

So rearranging to make E the subject I get:

$$E = \frac{17V}{5} - 24$$

So does V have to equal 12 to cancel out the +12V from the battery to make that branch have no p.d. along it?

Then that would mean E has to be 16.8V

Last edited: Nov 21, 2012
6. Nov 21, 2012

### Staff: Mentor

Something fishy with your equation; there's no "V" there, and the denominator of the first term is suspect.

However...
That's right.

7. Nov 21, 2012

### trollcast

I missed the V when I was typing the Latex bit. (fixed it now)

For the second part about the power in the 1R resistor do I need to work out the total resistance of the circuit, calculate the current draw from the voltage source and then use P = VI assuming all the current is delivered down the central branch?

8. Nov 21, 2012

### Staff: Mentor

The denominator value "2" is not correct.
When the charging current is zero the only current flowing will be in the right hand loop; the left hand loop will effectively not be there. So determine what that current will be given your calculated value for E and the resistances in the right hand loop. You can then work out the various powers.

9. Nov 21, 2012

### trollcast

So its just a series circuit with a Voltage supply of E volts, a 1R and a 0.4R resistor?

So the total resistance is 1.4 R

So by Ohms Law the current is 12A.

So the total power delivered is 201.6W.

and the power disipated in the 1R resistor is 144W

so the percentage is 71.4%?

For the last part do I need to rearrange my equation for the voltage from the top node to where the resistors join up so that V is the subject then input E = 20V?

Or is the voltage through the battery branch 8 as 20 - 12 = 8?

10. Nov 21, 2012

### Staff: Mentor

Correct.
Looks okay.
You'll need to determine the current in the battery branch with the new value of E in place.

One method is to do as you say and find the value of the node voltage V to begin with, then pull the battery branch's current term out of the node equation and find the current.

Note that when E = 20 Volts, the node voltage V won't be 20 V.

11. Nov 21, 2012

### trollcast

So rearrange the node equation to get

$$V = \frac{5E + 120}{17}$$

so V = 12.9412...

so the battery branches current term is,

$$I_{batt} = 5V - 60$$

so

$$I_{batt} = 4.706...$$

12. Nov 21, 2012

### Staff: Mentor

That looks fine.

13. Nov 21, 2012

### trollcast

Thanks

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