Question

A battery of \( 6 \, \text{V} \) is connected to the circuit as shown below. The current \( I \) drawn from the battery is: \begin{center} \includegraphics[width=0.4\textwidth]{50.png} % Replace with your image file \end{center}

• A. \( 1 \, \text{A} \)
• B. \( 2 \, \text{A} \)
• C. \( \frac{6}{11} \, \text{A} \)
• D. \( \frac{4}{3} \, \text{A} \)

Hint:

In a balanced Wheatstone bridge, the current through the middle resistor is zero. Use this property to simplify the circuit by removing that resistor.

Answer 1

The Correct Option is A

In a balanced Wheatstone bridge, the current through the middle resistor is zero, meaning that the \( 5 \, \Omega \) resistor does not contribute to the current. We can simplify the circuit by removing this resistor. Step 1: Simplify the circuit
Removeing the \( 5 \, \Omega \) resistor.
Combining the remaining resistors.
The \( 3 \, \Omega \) resistors in the top branch are in series, so their combined resistance is: \[ R_{\text{top}} = 3 + 3 = 6 \, \Omega. \] The \( 6 \, \Omega \) resistors in the bottom branch are in parallel, so their equivalent resistance is: \[ R_{\text{bottom}} = \frac{6 \cdot 6}{6 + 6} = \frac{36}{12} = 3 \, \Omega. \] Now, the two resistances \( R_{\text{top}} \) and \( R_{\text{bottom}} \) are in parallel: \[ R_{\text{eq}} = \frac{6 \cdot 3}{6 + 3} = \frac{18}{9} = 2 \, \Omega. \] Finally, add the \( 2 \, \Omega \) resistor in series with the equivalent resistance: \[ R_{\text{total}} = R_{\text{eq}} + 2 = 2 + 2 = 6 \, \Omega. \] Step 2: Apply Ohm's Law
The total current is given by: \[ I = \frac{V}{R_{\text{total}}}. \] Substitute \( V = 6 \, \text{V} \) and \( R_{\text{total}} = 6 \, \Omega \): \[ I = \frac{6}{6} = 1 \, \text{A}. \] Final Answer: \[ \boxed{1 \, \text{A}}. \]