Question

The bridge circuit, shown in Figure (a), can be equivalently represented using the circuit shown in Figure (b). The values of $R_1$, $R_2$, and $V_C$ in the equivalent circuit are

 

• A. $R_1 = 6 k\Omega$, $R_2 = 3 k\Omega$, and $V_C = 9 V$
• B. $R_1 = 3 k\Omega$, $R_2 = 6 k\Omega$, and $V_C = 4.5 V$
• C. $R_1 = 2 k\Omega$, $R_2 = 2 k\Omega$, and $V_C = 9 V$
• D. $R_1 = 2 k\Omega$, $R_2 = 2 k\Omega$, and $V_C = 4.5 V$

Hint:

Use Thevenin's theorem to simplify each branch of the bridge circuit. The common voltage source in the equivalent circuit can often be chosen as a reference point within the original source voltage.

Answer 1

The Correct Option is C

Step 1: Analyze the top branch to find \(V_H\) and \(R_{TH1}\).
\(V_H = 18 V \times \frac{6 k\Omega}{3 k\Omega + 6 k\Omega} = 12 V\).
\(R_{TH1} = R_1 = \frac{3 k\Omega \times 6 k\Omega}{3 k\Omega + 6 k\Omega} = 2 k\Omega\).
Step 2: Analyze the bottom branch to find \(V_L\) and \(R_{TH2}\).
\(V_L = 18 V \times \frac{3 k\Omega}{6 k\Omega + 3 k\Omega} = 6 V\).
\(R_{TH2} = R_2 = \frac{6 k\Omega \times 3 k\Omega}{6 k\Omega + 3 k\Omega} = 2 k\Omega\).
Step 3: Determine \(V_C\) by considering the midpoint of the source.
Let the ground be at 0 V. The top of the 18 V source is at 18 V. The midpoint is at 9 V. If we consider \(V_C = 9 V\), then \(V_H = V_C + V_1 \implies 12 V = 9 V + V_1 \implies V_1 = 3 V\). And \(V_L = V_C - V_2 \implies 6 V = 9 V - V_2 \implies V_2 = 3 V\) (with the polarity shown).
Step 4: Match the values with the options.
We found \(R_1 = 2 k\Omega\), \(R_2 = 2 k\Omega\), and \(V_C = 9 V\), which corresponds to option (C).