Calculating Current, Voltage in Series Circuit with 25V Battery and 2 Resistors

In summary, the conversation discusses a series circuit with a battery of 25 Volts and 2 resistors, one with 250 Ohms and the other with 500 Ohms. The desired values are the current, V1, and V2. Using Ohm's Law, the current is found to be 33.3 mA. However, due to rounding errors, the total voltage only adds up to 24.75 Volts instead of 25 Volts. The solution is to use more significant digits in the calculations to achieve a more accurate result.
  • #1
Shakerhood
9
0
I have a series Circuit with a battery of 25 Volts and 2 Resistors, 1 is 250 Ohms and the other is 500 Ohms. What is wanted is

a. I = _______mA
b. V1 = _______
c. V2 = _______

To get I I took I = V 25 25
______ ______ _____ = .033mA
R1 + R2 250 + 500 750

The Current stays constant in a series circuit, so for V1 and V2 I got 16.5 and 8.25, and that only totals 24.75V, and the battery Voltage is 25, so what am I doing wrong?
 
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  • #2
In series, the net resistance is additive, then use Ohm's Law to find the current. The sum of the voltage drops throughout the circuit must equal the voltage in the battery.

The current is constant in any circuit via conservation of charge arguments.

The discrepency is due to your rounding of the current.
 
  • #3
I don't understand because I took 25 Volts and divided it by 750 which is the sum of R1 and R2 and the answer is .033333333333333 ect. the answer was asked for in mA so I put .033mA. and like I said it comes to 24.75 Volts, so how can I fix this to keep current constant and my Voltage total to = 25?
 
  • #4
Your unit is wrong. Volt/ohm = amp (A). So the answer is 25/750=0.033333... A=33.3 mA
The other thing is rounding error. If you calculate with two significant digits the result will not be more accurate than two digits. That 24.75 rounded to two digits is just 25. The resistance data were given with 3 significant digits, you should give the result also with 3 digits. 750 (ohm) * 0.0333 (A) = 24.975 V. rounded to 3 digits it is 25.0 V.

ehild
 
  • #5
Thank you, I realized my first error on the mA's and then I just left the 3s go on the calculator and I was able to get it to equal 25 Volts!
 
  • #6
Shakerhood said:
Thank you, I realized my first error on the mA's and then I just left the 3s go on the calculator and I was able to get it to equal 25 Volts!

Like I said, the discrepency is due to the rounding in your current.
 

1. How do I calculate the current in a series circuit with a 25V battery and 2 resistors?

To calculate the current in a series circuit, you will need to use Ohm's Law, which states that current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the voltage is 25V and the resistance is the sum of the two resistors. Simply add the two resistances together and then divide the voltage by that sum to find the current.

2. What is the formula for calculating voltage in a series circuit with a 25V battery and 2 resistors?

The formula for calculating voltage in a series circuit is V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, the current is the same throughout the circuit, so you can use the same value for I when calculating the voltage across each resistor.

3. Can the current in a series circuit be greater than the battery voltage?

No, the current in a series circuit cannot be greater than the battery voltage. This is because the voltage is divided among the resistors in the circuit, resulting in a lower current. Additionally, the resistance of the circuit also limits the amount of current that can flow.

4. How do I calculate the total resistance in a series circuit with 2 resistors?

To calculate the total resistance in a series circuit, simply add the resistances of each individual resistor. So if you have 2 resistors with values of 10 ohms and 20 ohms, the total resistance would be 10 ohms + 20 ohms = 30 ohms.

5. What happens to the current if one of the resistors in a series circuit is removed?

If one of the resistors in a series circuit is removed, the total resistance of the circuit decreases. This means that the current in the circuit will increase. However, the current will still be limited by the battery voltage and the remaining resistance in the circuit.

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