How Do You Calculate Current and Terminal Voltage in a Mixed Battery Circuit?

[qety6321]

Homework Statement

two batteries A and B are connected in pareallel. A has emf of 108V and internal resistance of 3 Ω while B has emf of 120V and internal resistance of 2Ω. The battery terminal is connected to battery C in series with a 25 resistor. Battery C has emf of 30V and negligible internal resistance.

Homework Equations

find the value and direction of current in each battery and terminal voltage of B.

The Attempt at a Solution

I'm struggling to find the current in each battery first and I have tried to find the terminal voltage of B by using Kirchhoff's law and nodes but cannot get the right answer.

 

[NascentOxygen]

Hi qety6321. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Please attach your sketch of the circuit showing all the information given. Then show your calculations so we can see where you are going wrong.

 

[berkeman]

Homework Statement

two batteries A and B are connected in pareallel. A has emf of 108V and internal resistance of 3 Ω while B has emf of 120V and internal resistance of 2Ω. The battery terminal is connected to battery C in series with a 25 resistor. Battery C has emf of 30V and negligible internal resistance.

Homework Equations

find the value and direction of current in each battery and terminal voltage of B.

The Attempt at a Solution

I'm struggling to find the current in each battery first and I have tried to find the terminal voltage of B by using Kirchhoff's law and nodes but cannot get the right answer.

As a bonus question, calculate how long before one of the batteries catches fire...

 

[Chronum]

Try drawing the EMF ups and downs separately for the batteries. Something like a ----EMF---/\/\/---- diagram. Then use appropriate math.

 

[HalfLostAndSearching]

I would approach this problem by first calculating the total equivalent resistance of the circuit. In this case, it would be (1/3 + 1/2 + 1/25)^-1 = 5/6 Ω. This means that the total current flowing through the circuit is 30V / (5/6 Ω) = 36A.

Next, I would use Ohm's law to calculate the current in each battery. For battery A, the current would be (108V - 36A * 3 Ω) = 12A. For battery B, the current would be (120V - 36A * 2 Ω) = 12A.

To find the terminal voltage of battery B, I would use Kirchhoff's voltage law and set up an equation: 30V - 12A * 25 Ω - (120V - 36A * 2 Ω) = 0. Solving for the terminal voltage of B, I get 120V. This means that the direction of current in battery B is from positive to negative, as expected.

In summary, the current in each battery is 12A, and the terminal voltage of battery B is 120V. This means that battery B is supplying more current to the circuit than battery A, which is expected due to its higher emf.