Batteries in Parallel and series

In summary, the conversation discusses two circuits with two batteries of emf "E" and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor. The goal is to calculate the power dissipated by the resistor of resistance R2 in circuit A, given that EMF=10, R1=300 ohms, and R2=5000 Ohms. Using the equation P=V^2/R, the power is calculated to be 0.07142857143. However, the correct answer to 2 significant figures is 0.064. After some clarification, it is determined that the correct answer only
  • #1
Talz1994
6
0

Homework Statement


You are given two circuits with two batteries of emf "E" and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor.
Note that the symbol E should be entered in your answers as EMF.

What is the power dissipated by the resistor of resistance R2 for circuit A, given that
EMF=10 R1=300 ohms R2=5000 Ohms

Calculate the power to two significant figures.


Homework Equations


P=I^2.R
P=V^2/R


The Attempt at a Solution



well i used the second equation, and from previous question worked out my current in algebraic equation,
I=2E/2R1+R2 so I= 2x10/2x300+5000 =1/280 R=5000+2x300= 5600

P=I^2.R

P=(1/280)^2.5600= 0.07142857143

how ever the correct answer to 2.s.f is 0.064
 
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  • #2
Talz1994 said:

Homework Statement


You are given two circuits with two batteries of emf "E" and internal resistance R1 each. Circuit A has the batteries connected in series with a resistor of resistance R2, and circuit B has the batteries connected in parallel to an equivalent resistor.
Note that the symbol E should be entered in your answers as EMF.

What is the power dissipated by the resistor of resistance R2 for circuit A, given that
EMF=10 R1=300 ohms R2=5000 Ohms

Calculate the power to two significant figures.


Homework Equations


P=I^2.R
P=V^2/R


The Attempt at a Solution



well i used the second equation, and from previous question worked out my current in algebraic equation,
I=2E/2R1+R2 so I= 2x10/2x300+5000 =1/280 R=5000+2x300= 5600

P=I^2.R

P=(1/280)^2.5600= 0.07142857143

how ever the correct answer to 2.s.f is 0.064

They're only looking for the power dissipated by R2. You've included all the resistance in your R.
 
  • #3
oh ok thanks, thread can be closed now.
 

Related to Batteries in Parallel and series

1. What is the difference between batteries in parallel and series?

When batteries are connected in parallel, the positive terminals are connected together and the negative terminals are connected together. This increases the overall capacity of the batteries, meaning they can provide more energy for longer periods of time. In contrast, when batteries are connected in series, the positive terminal of one battery is connected to the negative terminal of the next battery. This increases the overall voltage of the batteries, meaning they can provide more power.

2. Can I mix batteries in parallel and series?

It is not recommended to mix batteries in parallel and series. This can lead to imbalanced charging and discharging, which can reduce the lifespan of the batteries and potentially cause safety hazards. It is best to use batteries of the same type, brand, and age when connecting them in parallel or series.

3. How do I calculate the total voltage of batteries in series?

To calculate the total voltage of batteries connected in series, simply add the individual voltages of each battery. For example, if you have three 1.5V batteries connected in series, the total voltage would be 4.5V (1.5V + 1.5V + 1.5V = 4.5V).

4. What happens if one battery in a series or parallel connection fails?

If one battery in a series connection fails, the entire circuit will be broken and no power will be provided. However, in a parallel connection, the remaining batteries will continue to provide power, but the overall voltage and capacity will be reduced. It is important to regularly check and replace any faulty batteries to maintain the efficiency and safety of the circuit.

5. Which connection is better for increasing battery life?

It depends on the specific application and requirements. Parallel connections are better for increasing battery life, as they increase the overall capacity and can provide more energy for longer periods of time. However, series connections are better for increasing power, as they increase the overall voltage and can provide more power. It is important to carefully consider the needs of the circuit before determining the best connection for your batteries.

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