Question

Consider the circuit shown in the figure. The current $I$ flowing through the 10 $\Omega$ resistor is ________________.

• A. 1 A
• B. 0 A
• C. 0.1 A
• D. -0.1 A

Hint:

When two branches of a circuit are perfectly symmetrical, the midpoint potentials become equal, leading to zero current through the connecting resistor.

Answer 1

The Correct Option is B

The circuit consists of two identical voltage sources (3 V each) and two identical sets of series resistances on the left and right sides. Each side has:
Left branch: $3\ \text{V} \rightarrow 1\Omega \rightarrow 2\Omega$
Right branch: $3\ \text{V} \rightarrow 2\Omega \rightarrow 1\Omega$
Step 1: Compute equivalent resistance of each branch.
Left branch resistance: \[ R_L = 1 + 2 = 3\Omega \] Right branch resistance: \[ R_R = 2 + 1 = 3\Omega \]
Step 2: Compute the current supplied by each battery.
\[ I_L = \frac{3}{3} = 1\ \text{A}, \qquad I_R = \frac{3}{3} = 1\ \text{A} \]
Step 3: Determine potentials at the two ends of the 10 $\Omega$ resistor.
Both branches produce the same voltage drop across identical resistances, making the potential at both nodes equal.
Thus, the 10 $\Omega$ resistor has the same voltage at both ends, meaning: \[ V_{\text{left}} = V_{\text{right}}. \]
Hence, \[ I = \frac{V_{\text{left}} - V_{\text{right}}}{10\Omega} = 0. \]
Final Answer: 0 A