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. \( 7.5 \text{ V} \)
• 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

We are given a circuit and asked to find the value of voltage $V$ such that the current flowing through the 3 $\Omega$ resistor is 0. Let's label the nodes to apply nodal analysis or analyze the condition. Let the node between the 1$\Omega$ and 3$\Omega$ resistor be $A$. Let the node between the 3$\Omega$ and 5$\Omega$ resistor be $B$. The bottom wire can be considered the reference node (ground, 0V). The condition given is that the current through the 3 $\Omega$ resistor is 0. This means $I_{3\Omega} = 0$. If the current through the 3 $\Omega$ resistor is 0, then the voltage drop across the 3 $\Omega$ resistor is also 0 ($V = IR$). Therefore, the voltage at node A must be equal to the voltage at node B: $V_A = V_B$. Now, let's find $V_A$ and $V_B$ in terms of the sources and resistors. 
Analyzing Node A: The 5V source is connected to the 1$\Omega$ resistor and node A. The 2$\Omega$ resistor is connected between node A and ground (0V). If $I_{3\Omega} = 0$, then all the current from the 5V source that passes through 1$\Omega$ must go through the 2$\Omega$ resistor. The current through the 1$\Omega$ resistor is $I_{1\Omega} = \frac{5 - V_A}{1}$. The current through the 2$\Omega$ resistor is $I_{2\Omega} = \frac{V_A - 0}{2}$. Since $I_{3\Omega} = 0$, by KCL at node A: $I_{1\Omega} = I_{2\Omega}$. $\frac{5 - V_A}{1} = \frac{V_A}{2}$ $2(5 - V_A) = V_A$ $10 - 2V_A = V_A$ $10 = 3V_A$ $V_A = \frac{10}{3}$ V. 
Analyzing Node B: Since $I_{3\Omega} = 0$, we have $V_B = V_A = \frac{10}{3}$ V. The 5$\Omega$ resistor is connected between node B and the voltage source $V$. The 4$\Omega$ resistor is connected between node B and ground (0V). If $I_{3\Omega} = 0$, then all the current coming from the source $V$ that passes through 5$\Omega$ must go through the 4$\Omega$ resistor. The current through the 5$\Omega$ resistor is $I_{5\Omega} = \frac{V - V_B}{5}$. The current through the 4$\Omega$ resistor is $I_{4\Omega} = \frac{V_B - 0}{4}$. By KCL at node B: $I_{5\Omega} = I_{4\Omega}$. $\frac{V - V_B}{5} = \frac{V_B}{4}$ Substitute $V_B = \frac{10}{3}$: $\frac{V - \frac{10}{3}}{5} = \frac{\frac{10}{3}}{4}$ $\frac{V - \frac{10}{3}}{5} = \frac{10}{12}$ $\frac{V - \frac{10}{3}}{5} = \frac{5}{6}$ $V - \frac{10}{3} = 5 \times \frac{5}{6}$ $V - \frac{10}{3} = \frac{25}{6}$ $V = \frac{25}{6} + \frac{10}{3}$ To add these fractions, find a common denominator, which is 6: $V = \frac{25}{6} + \frac{10 \times 2}{3 \times 2}$ $V = \frac{25}{6} + \frac{20}{6}$ $V = \frac{25 + 20}{6}$ $V = \frac{45}{6}$ $V = \frac{15}{2}$ $V = 7.5$ V. 
Thus, the value of $V$ for which the current in the 3 $\Omega$ resistor is 0 is 7.5 V.