Question

Find the value of $V$ when the current in the 3 $\Omega$ resistor is 0. (Circuit diagram provided: A circuit with two voltage sources, 5V and V, and several resistors: 1$\Omega$, 2$\Omega$, 3$\Omega$, 4$\Omega$, 5$\Omega$. The 1$\Omega$ and 3$\Omega$ resistors are in the top branch. The 2$\Omega$ and 4$\Omega$ resistors are in parallel branches.)

• A. \( 3.5 \text{ V} \)
• B. \( 6.5 \text{ V} \)
• C. \( 7.1 \text{ V} \)
• D. \( 8.5 \text{ V} \)

Hint:

When a problem states that the current through a specific resistor is zero, it implies that there is no voltage drop across that resistor. This means the voltage at its two terminals is the same. This crucial insight simplifies the circuit analysis, often allowing you to treat the two connected nodes as a single node or to use voltage division principles. In this case, it effectively divides the problem into two independent voltage divider problems, one for each side of the zero-current resistor.

Answer 1

The Correct Option is C

1. Understanding the Problem:

- We are given a circuit with resistors of 1Ω, 2Ω, 3Ω, 4Ω, and 5Ω, and voltage sources of 5V and \( V \).
- The question asks to find the value of \( V \) when the current through the 3Ω resistor is 0. This implies that no current is flowing through the 3Ω resistor, meaning the circuit bypasses this resistor.

2. Analyzing the Circuit:

When the current through the 3Ω resistor is zero, it does not affect the circuit. The remaining resistors (1Ω, 2Ω, 4Ω, and 5Ω) will form a simplified circuit.
- The resistors 1Ω and 2Ω are in series, and the resistors 4Ω and 5Ω are in parallel.
- The total resistance of the 1Ω and 2Ω series combination is:

\[ R_{\text{series}} = 1\ \Omega + 2\ \Omega = 3\ \Omega \]

Next, we need to calculate the equivalent resistance of the parallel combination of the 4Ω and 5Ω resistors:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{4\ \Omega} + \frac{1}{5\ \Omega} = \frac{5 + 4}{20} = \frac{9}{20} \]

Thus, the equivalent parallel resistance is:

\[ R_{\text{parallel}} = \frac{20}{9}\ \Omega \]

The total resistance in the circuit is the sum of the series and parallel resistances:

\[ R_{\text{total}} = R_{\text{series}} + R_{\text{parallel}} = 3\ \Omega + \frac{20}{9}\ \Omega = \frac{27}{9}\ \Omega + \frac{20}{9}\ \Omega = \frac{47}{9}\ \Omega \]

3. Applying Kirchhoff's Voltage Law (KVL):

Now, applying Kirchhoff's Voltage Law to the loop with the total voltage \( V \) and the total resistance \( R_{\text{total}} \), the total current \( I \) in the circuit is:

\[ I = \frac{5\ \text{V}}{R_{\text{total}}} = \frac{5}{\frac{47}{9}} = \frac{5 \times 9}{47} = \frac{45}{47}\ \text{A} \]

Next, the potential difference across the parallel resistors (4Ω and 5Ω) is:

\[ V_{\text{parallel}} = I \times R_{\text{parallel}} = \frac{45}{47} \times \frac{20}{9} = \frac{900}{423} \approx 2.13\ \text{V} \]

4. Finding the Value of \( V \):

Now, to find the value of \( V \), we apply the following relationship across the circuit:

\[ V = 5\ \text{V} + V_{\text{parallel}} = 5\ \text{V} + 2.13\ \text{V} = 7.13\ \text{V} \]

Final Answer:

The value of \( V \) is approximately 7.13 V.