# Homework Help: Calculate resistance if the switch is closed 2 battery with different voltage

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1. Jun 22, 2017

### FJay

• Member advised to use the homework template for posts in the homework sections of PF.
1. The problem statement, all variables and given/known data
before that, can you tell me about "current flow in this circuit" if :
A. the switch is open
B. the switch is closed

the Question is
No. 1 Calculate the resistance of R if the value of the current I in the figure is 0.50 A. ( i think it is related to no 2)

no 2. Calculate the resistance of R if the switch S is closed and the value of the current in
the resistor R2 is 0.0 A. (what i want to know)

2. Relevant equations
Kirchoff's Law (I & II) (i think i missed something because if that i cant solve this)
V = I R

3. The attempt at a solution

my thoughts are:
1.) what i thought is because the switch is open, the current will ignore R2 and E2, and just go through R and R1(R and R1 will be series)
that will be ez to find out R for no 1.
V / I = R
18 / 0.5 = 12+R
R = 24

2) when switch is closed, it has 2 battery and its voltage will be 18V (if both of the batteries have same voltage) but in this case its have different voltage.
after i search it, someone tell battery with higher voltage will destroy battery with lower voltage. then the voltage for this circuit just 18V.
and because higher voltage will destroy the lower one, how current flow for this circuit.

and please correct me if im wrong

Last edited: Jun 22, 2017
2. Jun 22, 2017

### jbriggs444

You will need to give us your thoughts and show some work before we can help. The template is there for a reason.

3. Jun 22, 2017

### FJay

1.) what i thought is because the switch is open, the current will ignore R2 and E2, and just go through R and R1(R and R1 will be series)
that will be ez to find out R for no 1.
V / I = R
18 / 0.5 = 12+R
R = 24

2) when switch is closed, it has 2 battery and its voltage will be 18V (if both of the batteries have same voltage) but in this case its have different voltage.
after i search it, someone tell battery with higher voltage will destroy battery with lower voltage. then the voltage for this circuit just 18V.
and because higher voltage will destroy the lower one, how current flow for this circuit.

and please correct me if im wrong

4. Jun 22, 2017

### Staff: Mentor

That looks good.
No, the batteries are safe. There's resistors in the path of their currents which will limit the current flow and create potential drops between the batteries. It's only when (ideal) batteries of different voltages are directly connected in parallel that you need to worry about the current and power becoming unrealistic (heading to infinities).

Think about what conditions must exist for there to be no current drawn from E2 when its branch (E2 and R2 in series) is connected. What drives current in a circuit?

5. Jun 22, 2017

### FJay

I = V / R
if I = 0 with R = 4 then I = 0/4 thats mean on that R2 doesn't have more voltage (V=0) because all the voltage used by R3 ?

because the batteries have different voltage, the higher voltage will charge the lower one until they have same voltages. then Voltage on R3(after upper node), V = V1+V2 / 2 = 15V.

and before reach the upper node the V1 (left side battery) still have 18 Volt.

i think i need current flow illustration for this one :v
CMIIW.

is there any another formula ?

Last edited: Jun 22, 2017
6. Jun 22, 2017

### jbriggs444

There is no R3 on the drawing. The phrase "voltage used" is troubling. Voltage is not something that is used up. Instead, "voltage" is the potential difference between two points. If no current is flowing across R2, that means that the potential difference between its two terminals is zero.
If there is no current flowing across R2, how much current is flowing across the right hand battery?

7. Jun 22, 2017

### FJay

i get it now after searching some images.

because it has junctions we can Kirchhoffs Law I
I1 + I3 = I2, because I3 = 0 then I1=I2

using Kirchhoff's Law II find loop for both sides

E2 = I212 + 0*4 ---> 12 = 12I2 ---> I2 = 1

E1 = I1R + I212 ---> with I=1 ---> 18-12 = R, R=6 ohms