According to the Kirchoff's law of voltages:gneill said:
VT=2-3+5=4Vgneill said:
No. Check the polarities of the voltage sources as you go around the loop. Suppose you "walk" clockwise around the loop starting at the bottom of V1. Is there a potential rise or drop as you pass through each source? Do they have different polarity orientations with respect to your "walk"?saulwizard1 said:
The first one is=-2gneill said:
gneill said:
Okay, you should now be able to write out KVL for the loop, including the drops associated with the resistances.saulwizard1 said:
10V-(6 ohms*I)=0gneill said:
Ok, thank you so much for the help.gneill said:
The total voltage in a series circuit is the sum of all the individual voltage drops. In this circuit, the total voltage would be 2V + 3V + 5V = 10V.
The current in a series circuit is the same at all points. To calculate the current, you can use Ohm's Law: I = V/R. In this circuit, the current would be 10V / (4Ω + 2Ω) = 1.67A.
In a series circuit, the voltage drop across each resistor depends on its resistance. The voltage drop can be calculated using Ohm's Law: V = IR. In this circuit, the voltage drop across the 4Ω resistor would be 1.67A x 4Ω = 6.68V, and the voltage drop across the 2Ω resistor would be 1.67A x 2Ω = 3.34V.
In a series circuit, the total resistance is equal to the sum of the individual resistances. In this circuit, the total resistance would be 4Ω + 2Ω = 6Ω.
If another 2V battery was added to the circuit, the total voltage would increase to 12V, and the current would also increase. This is because the total resistance in the circuit remains the same, but the total voltage has increased, following Ohm's Law: I = V/R.
