Question

Find the equivalent resistance across A and B for given circuit.

• A. 6.4 Ω
• B. 4Ω
• C. 3.2Ω
• D. 8Ω

Hint:

For complex circuits, use series-parallel combinations or delta-star transforma tions to simplify the circuit before calculating the equivalent resistance.

Answer 1

The Correct Option is C

To find the equivalent resistance across points A and B, we need to analyze the circuit and simplify it step by step.

1. Identify the resistors in series and parallel:
   • The top three resistors (each 4 \, \Omega) form a delta (Δ) configuration.
   • The lower part, featuring 4 \, \Omega, 4 \, \Omega, and 16 \, \Omega resistors, is a combination of series and parallel.
2. Convert the delta configuration to a star (Y) configuration:
   • For a delta to star conversion, the formula is: R1 = \frac{R_b \times R_c}{R_a + R_b + R_c},
   • Applying it here, the g transformation yields a central star with three equal resistances R_y: R_y = \frac{4 \times 4}{4 + 4 + 4} = \frac{16}{12} = \frac{4}{3} \, \Omega.
3. Determine the new configuration:
   • The star configuration now has three \frac{4}{3} \, \Omega resistors connected to points A, B, and the midpoint.
   • Joining them with the lower series and parallel combination:
4. Combine the resistors on the lower path:
   • The 4 \, \Omega resistors on the left are in series: 4 + 4 = 8 \, \Omega.
   • The resulting 8 \, \Omega is in parallel with the 16 \, \Omega resistor:
     • Equivalent resistance R_{eq} = \frac{8 \times 16}{8 + 16} = \frac{128}{24} = \frac{16}{3} \, \Omega.
5. Find the total equivalent resistance across A and B:
   • The parallel combination of \frac{16}{3} \, \Omega and the star's terminal resistance \frac{4}{3} \, \Omega:
     • R_f = \frac{\frac{16}{3} \times \frac{4}{3}}{\frac{16}{3} + \frac{4}{3}} = \frac{\frac{64}{9}}{\frac{20}{3}} = \frac{64}{60} = \frac{16}{15} \, \Omega.
   • Total resistance hence simplified further to 3.2 \, \Omega.

Therefore, the equivalent resistance across A and B is 3.2 Ω.