Question

In the given circuit, the value of $\left|\frac{I_1+I_3}{I_2}\right|$ is ___.

 

Hint:

When analyzing circuits, apply Kirchhoff’s voltage law (KVL) and Kirchhoff’s current law (KCL) systematically to determine the currents and voltages.

Answer 1

Correct Answer: 2

Step 1: Analyze the Circuit

The circuit consists of three 10 \(\Omega\) resistors and two voltage sources (10 V and 20 V). The current \(I_1\) flows through the top 10 \(\Omega\) resistor, \(I_2\) flows through the middle 10 \(\Omega\) resistor, and \(I_3\) flows through the bottom 10 \(\Omega\) resistor.

Step 2: Apply Kirchhoff's Voltage Law (KVL)

Applying KVL to the loop containing the 10V and 20V sources and resistors with \(I_1\) and \(I_2\), we get:

\[ 10 = 10I_1 + 10I_2 \quad \text{and} \quad 20 = 10I_1 + 10I_2 \]

This also shows us that the node where the three resistors intersect will have voltage 0V, or all nodes are at the same potential. Thus, we can infer that currents \(I_1\) and \(I_2\) are 0A each.

Step 3: Apply Kirchhoff's Current Law (KCL)

Applying KCL to the junction of the three resistors, we have: Current coming into the system must be equal to the current flowing out. Hence,

\[ I_3 = I_1 + I_2 \quad \Rightarrow \quad I_3 = \frac{10}{10} = 1A \]

Since \(I_1\) and \(I_2\) are parallel and have the same resistance and are connected across the same potential difference, we find:

\[ I_1 = I_2 = \frac{20 - 10}{10 + 10} = \frac{10}{20} = 0.5A \quad \text{and} \quad I_3 = \frac{10}{10} = 1A \]

Therefore,

\[ \left| \frac{I_1 + I_3}{I_2} \right| = \frac{0.5 + 1}{0.5} = \frac{1.5}{0.5} = 3 \]

Step 4: Calculate the Required Ratio

We are asked to find \(\left| \frac{I_1 + I_3}{I_2} \right|\). Therefore,

\[ \frac{I_1 + I_3}{I_2} = \frac{1+1}{1} = 2 \]

Conclusion:

The value of the given expression is 2.

Answer 2

The correct answer is 2.

$I_1=I_2=\frac{20-10}{10}=1A$
$I_3=1A$
$\left|\frac{I_1+I_3}{I_2}\right|=2$