Can somebody check my answers (DC-circuit)

  • Thread starter cseet
  • Start date
In summary, the conversation is about finding the terminal voltage and resistance of a circuit powered by two 1.55 V batteries connected in series. It also discusses how to calculate the fraction of power dissipated in the batteries. The solution involves using Ohm's law and the power formula.
  • #1
cseet
40
0
Hi all

need somebody to go thru my answers are actually correct for DC circuits...

thanks in advance...

question:
Two 1.55 V batteries are connected in series to power a lamp. One battery has an internal resistance r1 = 0.208 ohm, and the other has r2 = 0.169 ohm. When the switch is closed a current of 0.55 A flows and the lamp lights up.

answer:
What is the terminal voltage of the battery combination and what is the resistance R of the lamp filament?

question:
What fraction of power is dissipated in the batteries?
----------------------------------------------------------------------------------

can somebody advise me on how I get the fraction of power answer?
my conclusion was (correct me if I was wrong):

Power = V I, so

power in internal r Vi I
------------------------ = ------ = am I in the right track?
power in whole circuit V I

thanks again
cseet
 
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  • #2
cseet said:
can somebody advise me on how I get the fraction of power answer?
my conclusion was (correct me if I was wrong):
The total power delivered by the batteries is given by P = VI, where V is the total voltage of the two batteries. This power is dissipated in the resistance of the circuit, some of which is internal to the batteries. The power dissipated in a resistor is P = VRI, where VR is the voltage drop across the resistor. From Ohm's law, the voltage drop across a resistor is V = IR, so the power dissipated in a reistor becomes: P = I2R.

So... Find the power dissipated in the internal resistance of the batteries and divide by the total power. So you are on the right track if by Vi you meant the voltage drop due to the internal resistance.
 
  • #3
By the way, if I'm not mistaken the fraction of power that is dissipated outside the batteries is called the circuit's efficiency (I think it's denoted by the letter [tex]\mu[/tex]).
 

1. What is a DC-circuit?

A DC-circuit refers to a direct current circuit, which is an electrical circuit that allows current to flow in only one direction. This is in contrast to an alternating current (AC) circuit, which allows current to flow in both directions.

2. How do I check my answers for a DC-circuit?

To check your answers for a DC-circuit, you can use Kirchhoff's laws and Ohm's law. Kirchhoff's laws state that the sum of currents entering a node must equal the sum of currents leaving the node, and the sum of voltage drops in a closed loop must equal the sum of voltage sources. Ohm's law states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance.

3. What is the purpose of checking my answers for a DC-circuit?

The purpose of checking your answers for a DC-circuit is to ensure that your calculations are accurate and that the circuit will function properly. It also helps to identify any errors or mistakes in your calculations, which can be corrected to improve the overall performance of the circuit.

4. Can I use a calculator to check my answers for a DC-circuit?

Yes, you can use a calculator to check your answers for a DC-circuit. In fact, a calculator can be very helpful in performing complex calculations involving resistors, voltage sources, and current flows. It is important to use the correct formulas and units when using a calculator to check your answers.

5. Are there any tips for checking my answers for a DC-circuit?

Some tips for checking your answers for a DC-circuit include using a systematic approach, double-checking your calculations, and checking for common mistakes such as incorrect unit conversions or using the wrong formula. It can also be helpful to work through the problem using different methods to verify your results.

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