Question

Find $R_{AB}$.

Answer 1

3R

Answer 2

To find the equivalent resistance $R_{AB}$ for the given circuit, we first analyze the circuit structure and identify any open ends.

1. Identify Open Ends: The problem asks for $R_{AB}$, implying that current enters at A and exits at B. Points C and D are the ends of the top and bottom branches, respectively. Since nothing is connected to C and D, they are considered open circuits. This means no current flows through the resistor connected between Q and C, and no current flows through the resistor connected between D and S. Therefore, these two resistors can be removed from the circuit for the calculation of $R_{AB}$.

2. Redraw the Simplified Circuit: After removing the resistors R(Q-C) and R(D-S), the circuit simplifies to:

3. Identify Series/Parallel Combinations: The circuit can be viewed as three parts in series:

   • Resistor between A and P ($R_{AP} = R$).
   • A bridge-like network between P and T.
   • Resistor between T and B ($R_{TB} = R$).

4. Simplify the Bridge Network (P-Q-S-T): Consider the section of the circuit between points P and T. This section consists of four resistors: R(P-Q), R(P-S), R(Q-T), and R(S-T). We can see two parallel paths from P to T:

   • Path 1: P → Q → T. The total resistance of this path is $R_{PQT} = R_{PQ} + R_{QT} = R + R = 2R$.
   • Path 2: P → S → T. The total resistance of this path is $R_{PST} = R_{PS} + R_{ST} = R + R = 2R$.

   Since these two paths are in parallel between P and T, their equivalent resistance ($R_{PT}$) is: $R_{PT} = \frac{R_{PQT} \times R_{PST}}{R_{PQT} + R_{PST}} = \frac{2R \times 2R}{2R + 2R} = \frac{4R^2}{4R} = R$

5. Calculate Total Equivalent Resistance $R_{AB}$: Now, the entire circuit is a series combination of $R_{AP}$, $R_{PT}$, and $R_{TB}$: $R_{AB} = R_{AP} + R_{PT} + R_{TB}$ $R_{AB} = R + R + R$ $R_{AB} = 3R$

The final answer is $\boxed{3R}$.

Explanation of the solution:

1. Identify open ends C and D; remove resistors R(Q-C) and R(D-S).
2. The circuit simplifies to A-R-P, a bridge network P-T, and T-R-B.
3. The bridge network (P-Q-T and P-S-T) consists of two parallel branches, each with resistance 2R. Their equivalent resistance is $R_{PT} = (2R \times 2R) / (2R + 2R) = R$.
4. The total equivalent resistance $R_{AB}$ is the series combination of R(A-P), $R_{PT}$, and R(T-B), which is $R + R + R = 3R$.

Answer: $3R$