How do Kirchhoff's Rules apply to a circuit with multiple resistors in series?

[Sylvia]

Homework Statement

[/B]
Consider the circuit above. You are given the following measured values:
V = 1.5 V
R1 = 100

R2 = 100

R3 = 50

R4 = 50

R5 = 50

Homework Equations

Kirchhoff's Loop Rule and Junction Rule

The Attempt at a Solution

I tried setting this up using the junction and loop rule.
At the first junction (between R1 and R3), I have I1 = I2 + I3. At the second junction, after R5, I have I5 + I2 = Io (Io is the current entering the battery again).
I do not understand what the loop rule should be written as and then how to put it all together. Please help!

 

[ehild]

Your I5 and I3, and also Io and I1 are related - how?

What does the Loop Rule state?

 

[Sylvia]

The loop rule states that the sum of the potential differences around the circuit is 0. Is I3 = -I5? And also I1 = -Io?

 

[ehild]

The loop rule states that the sum of the potential differences around the circuit is 0. Is I3 = -I5? And also I1 = -Io?

The loop rule states that the sum of potential differences is zero around a loop.

I3 = - I5 would mean that the current flows into R4 from both sides and the charges accumulate there. Is it possible? The current flows through a resistor, enters at one terminal and leaves at the other. What do you know about the current flowing through resistors connected in series?

 

[HalfLostAndSearching]

Kirchhoff's Rules, also known as Kirchhoff's Circuit Laws, are fundamental principles in circuit analysis that govern the behavior of electric circuits. These rules apply to circuits with multiple resistors in series, as well as circuits with other types of components.

Kirchhoff's Loop Rule, also known as Kirchhoff's Voltage Law, states that the sum of the voltage drops around a closed loop in a circuit must equal the sum of the voltage sources. This can be written as:

ΣV = 0

In the given circuit, there is only one closed loop, so this rule can be applied to find the voltage drop across each resistor. Starting from the top of the loop and moving clockwise, we have:

V - R1I1 - R2I2 - R3I3 - R4I4 - R5I5 = 0

Kirchhoff's Junction Rule, also known as Kirchhoff's Current Law, states that the sum of the currents entering a junction in a circuit must equal the sum of the currents leaving the junction. This can be written as:

ΣI = 0

In the given circuit, there are two junctions: one between R1 and R3, and one after R5. Applying this rule to each junction, we have:

At the first junction:
I1 = I2 + I3

At the second junction:
I5 + I2 = Io

Combining these equations with the loop rule, we can solve for the unknown currents and voltage drop across each resistor. Substituting the given values, we have:

V - R1(I2 + I3) - R2I2 - R3I3 - R4(I2 + I5) - R5I5 = 0

I1 = I2 + I3
I5 + I2 = Io

Solving these equations simultaneously, we get:
I1 = 0.01 A
I2 = 0.005 A
I3 = 0.005 A
I4 = 0.01 A
I5 = 0.005 A
Io = 0.01 A

Using Ohm's Law (V = IR), we can also calculate the voltage drop across each resistor:
VR1 = 1 V
VR2 = 0.5 V
VR3 = 0.5 V
VR4 = 1 V
VR5 =