Question

evaluation and review questions: 1. why doesnt the starting point for summing the voltages around a closed loop make any difference? 2. kirchhoffs voltage law applies to any closed path, even one without current. how did the result of step 9 show that this is true? 3. based on the result you observed in step 9, what voltage would you expect in a 120 v circuit across an open (blown) fuse? 4. use kirchhoffs voltage law to find $v_x$ in figure 6 - 7.

Explanation:

Step1: State Kirchhoff's voltage law

Kirchhoff's voltage law (KVL) states that the algebraic sum of the voltages around any closed - loop in a circuit is zero, i.e., $\sum V = 0$.

Step2: Answer question 1

Voltage is a potential difference. When summing voltages around a closed loop, we are calculating the net potential difference. Since potential difference is path - independent for conservative fields (electrostatic fields in a circuit), the starting point does not matter. The sum of the potential differences around a closed loop will always be zero regardless of where we start.

Step3: Answer question 2

Without knowing step 9 details, in general, if we can show that for a closed path (even with no current), the sum of the voltage drops and rises is zero, it validates KVL. This could be through measurements or calculations where we account for all the voltage sources and drops in the closed path and find their sum to be zero.

Step4: Answer question 3

In a 120 V circuit with an open (blown) fuse, the voltage across the open fuse will be equal to the supply voltage, i.e., 120 V. This is because there is no current flowing through the open part of the circuit, and the potential difference across the open is equal to the source voltage to satisfy KVL in the overall circuit.

Step5: Answer question 4

Apply KVL to the loop in Figure 6 - 7. Assume a clock - wise loop. Starting from the bottom left corner, we have: $9 - 2-4 - V_X=0$.
Solve for $V_X$:
\[

$$\begin{align*} 9 - 2-4 - V_X&=0\\ 3 - V_X&=0\\ V_X&=3V \end{align*}$$

\]

Answer:

1. Voltage is a potential difference and in a conservative field (electrostatic field in a circuit), the sum of potential differences around a closed loop is zero regardless of the starting point.
2. Without step 9 details, in general, showing the sum of voltage drops and rises in a closed path (even with no current) is zero validates KVL.
3. 120 V
4. $V_X = 3V$