Question

In the circuit shown in the figure,

$R_{AB}=?$

• A. $\frac{21}{4}$ $\Omega$
• B. $\frac{5}{6}$ $\Omega$
• C. 10 $\Omega$
• D. 8 $\Omega$

Answer 1

Answer: (A)

$\frac{21}{4}$ $\Omega$

Answer 2

1. Identify the circuit type: The circuit contains a bridge-like structure CDEF with two diagonals $R_{CF}$ (6Ω) and $R_{DE}$ (6Ω).

2. Check for balanced Wheatstone bridge: Consider the bridge C-D-F-E with $R_{DE}$ as the central resistor. The arms are $R_{CD}=2\Omega$, $R_{DF}=6\Omega$, $R_{CE}=2\Omega$, $R_{EF}=6\Omega$.

   The condition for a balanced bridge (no current through $R_{DE}$) is $\frac{R_{CD}}{R_{DF}} = \frac{R_{CE}}{R_{EF}}$.

   Substituting the values: $\frac{2}{6} = \frac{2}{6}$, which simplifies to $\frac{1}{3} = \frac{1}{3}$.

   Since the condition is met, the bridge is balanced. This means the potential at node D is equal to the potential at node E ($V_D = V_E$).

3. Simplify the circuit by removing the balanced arm and merging nodes: Since $V_D = V_E$, no current flows through the 6Ω resistor between D and E. This resistor can be removed. Furthermore, since D and E are at the same potential, they can be merged into a single node, let's call it D'.

4. Redraw the simplified circuit and calculate equivalent resistances:

   • The 2Ω resistor ($R_{CD}$) and 2Ω resistor ($R_{CE}$) are now in parallel between C and D'.

     $R_{CD'} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega$.

   • The 6Ω resistor ($R_{DF}$) and 6Ω resistor ($R_{EF}$) are now in parallel between D' and F.

     $R_{D'F} = \frac{6 \times 6}{6 + 6} = 3 \, \Omega$.

   • The 6Ω resistor ($R_{CF}$) remains connected between C and F.

   The circuit now consists of:

   • $R_{AC} = 2 \, \Omega$
   • $R_{BD'} = 2 \, \Omega$
   • $R_{CD'} = 1 \, \Omega$
   • $R_{D'F} = 3 \, \Omega$
   • $R_{CF} = 6 \, \Omega$

5. Apply Nodal Analysis: Let $V_B = 0$ and $V_A = V$. Let $V_C, V_{D'}, V_F$ be the potentials at nodes C, D', F respectively.

   • Node C: $\frac{V_A - V_C}{2} = \frac{V_C - V_{D'}}{1} + \frac{V_C - V_F}{6}$

     Multiplying by 6: $3(V_A - V_C) = 6(V_C - V_{D'}) + (V_C - V_F)$

     $3V_A = 10V_C - 6V_{D'} - V_F$ (Eq. 1)

   • Node D': $\frac{V_C - V_{D'}}{1} = \frac{V_{D'} - V_B}{2} + \frac{V_{D'} - V_F}{3}$

     Multiplying by 6: $6(V_C - V_{D'}) = 3(V_{D'} - 0) + 2(V_{D'} - V_F)$

     $6V_C = 11V_{D'} - 2V_F$ (Eq. 2)

   • Node F: $\frac{V_C - V_F}{6} + \frac{V_{D'} - V_F}{3} = 0$

     Multiplying by 6: $(V_C - V_F) + 2(V_{D'} - V_F) = 0$

     $V_C + 2V_{D'} = 3V_F$ (Eq. 3)

6. Solve the system of equations:

   From Eq. 3: $V_F = \frac{V_C + 2V_{D'}}{3}$

   Substitute $V_F$ into Eq. 2:

   $6V_C = 11V_{D'} - 2\left(\frac{V_C + 2V_{D'}}{3}\right)$

   $18V_C = 33V_{D'} - 2V_C - 4V_{D'}$

   $20V_C = 29V_{D'} \implies V_{D'} = \frac{20}{29}V_C$

   Substitute $V_{D'}$ into $V_F$:

   $V_F = \frac{V_C + 2\left(\frac{20}{29}V_C\right)}{3} = \frac{V_C\left(1 + \frac{40}{29}\right)}{3} = \frac{V_C\left(\frac{69}{29}\right)}{3} = \frac{23}{29}V_C$

   Substitute $V_{D'}$ and $V_F$ into Eq. 1:

   $3V_A = 10V_C - 6\left(\frac{20}{29}V_C\right) - \frac{23}{29}V_C$

   $3V_A = V_C\left(10 - \frac{120}{29} - \frac{23}{29}\right) = V_C\left(\frac{290 - 120 - 23}{29}\right) = V_C\left(\frac{147}{29}\right)$

   $V_C = \frac{3 \times 29}{147}V_A = \frac{87}{147}V_A = \frac{29}{49}V_A$

7. Calculate the total current and equivalent resistance:

   The total current flowing from A to B is $I_{AB} = I_{AC} = \frac{V_A - V_C}{2}$.

   $I_{AB} = \frac{V_A - \frac{29}{49}V_A}{2} = \frac{V_A\left(\frac{49 - 29}{49}\right)}{2} = \frac{V_A\left(\frac{20}{49}\right)}{2} = \frac{10V_A}{49}$.

   The equivalent resistance $R_{AB} = \frac{V_A - V_B}{I_{AB}} = \frac{V_A - 0}{I_{AB}} = \frac{V_A}{\frac{10V_A}{49}} = \frac{49}{10} = 4.9 \, \Omega$.

8. Compare with options:

   (A) $\frac{21}{4} = 5.25 \, \Omega$ (B) $\frac{5}{6} \approx 0.83 \, \Omega$ (C) $10 \, \Omega$ (D) $8 \, \Omega$

   The calculated value $4.9 \, \Omega$ is closest to option (A) $5.25 \, \Omega$.

The final answer is $\boxed{\text{A}}$