The effective resistance between the points A and B of the circuit shown in the figure is:
Show Hint
Whenever a Wheatstone bridge is not balanced, direct series-parallel reduction generally fails. In such cases, the Node Voltage Method is often the fastest and most reliable technique. Assume a voltage across the terminals, determine the current drawn, and then use
\[
R_{\text{eq}}=\frac{V}{I}.
\]
This approach works for every linear resistor network.
Concept:
The given network is an unbalanced Wheatstone-bridge type resistor arrangement. Since the bridge is not balanced, the central resistor cannot be ignored and simple series-parallel reduction is not possible.
In such situations, the most systematic approach is the Node Voltage Method. We assume a potential difference across terminals A and B and determine the current drawn from the source. Once the total current is known, the equivalent resistance can be calculated using Ohm's law:
\[
R_{AB}=\frac{V}{I}.
\]
Let the junction of the upper branch be denoted by $C$ and the junction of the lower branch be denoted by $D$.
The resistor values are:
\[
R_{AC}=10\,\Omega,
\]
\[
R_{AD}=20\,\Omega,
\]
\[
R_{CB}=20\,\Omega,
\]
\[
R_{DB}=10\,\Omega,
\]
and the bridge resistor
\[
R_{CD}=10\,\Omega.
\]
Step 1: Check whether the Wheatstone bridge is balanced or not.
For a balanced Wheatstone bridge,
\[
\frac{R_{AC}}{R_{AD}}
=
\frac{R_{CB}}{R_{DB}}.
\]
Substituting the given values,
\[
\frac{10}{20}
=
\frac{1}{2}
\]
whereas
\[
\frac{20}{10}
=
2.
\]
Since
\[
\frac{1}{2}\neq 2,
\]
the bridge is not balanced.
Therefore, current will flow through the central resistor and we cannot remove it from the circuit.
Hence, nodal analysis must be used.
Step 2: Assign potentials to the nodes.
Let
\[
V_A=V
\]
and
\[
V_B=0.
\]
Assume the potentials at the intermediate nodes are
\[
V_C
\]
and
\[
V_D.
\]
Our objective is to determine $V_C$ and $V_D$.
Step 3: Apply Kirchhoff's Current Law (KCL) at node $C$.
The algebraic sum of currents leaving node $C$ must be zero.
Therefore,
\[
\frac{V_C-V}{10}
+
\frac{V_C-V_D}{10}
+
\frac{V_C-0}{20}
=
0.
\]
Multiplying throughout by $20$,
\[
2(V_C-V)
+
2(V_C-V_D)
+
V_C
=
0.
\]
Expanding,
\[
2V_C-2V
+
2V_C-2V_D
+
V_C
=
0.
\]
Combining like terms,
\[
5V_C-2V_D
=
2V.
\]
Thus,
\[
5V_C-2V_D=2V.
\]
\[
\boxed{5V_C-2V_D=2V}
\]
Step 4: Apply Kirchhoff's Current Law (KCL) at node $D$.
Similarly,
\[
\frac{V_D-V}{20}
+
\frac{V_D-V_C}{10}
+
\frac{V_D}{10}
=
0.
\]
Multiplying throughout by $20$,
\[
(V_D-V)
+
2(V_D-V_C)
+
2V_D
=
0.
\]
Expanding,
\[
V_D-V
+
2V_D-2V_C
+
2V_D
=
0.
\]
Collecting terms,
\[
-2V_C+5V_D
=
V.
\]
Hence,
\[
\boxed{-2V_C+5V_D=V}
\]
Step 5: Solve the simultaneous equations.
We have
\[
5V_C-2V_D=2V
\]
and
\[
-2V_C+5V_D=V.
\]
Multiplying the first equation by $5$,
\[
25V_C-10V_D=10V.
\]
Multiplying the second equation by $2$,
\[
-4V_C+10V_D=2V.
\]
Adding,
\[
21V_C=12V.
\]
Therefore,
\[
V_C=\frac{12}{21}V
=
\frac{4}{7}V.
\]
Substituting into
\[
-2V_C+5V_D=V,
\]
we get
\[
-2\left(\frac{4}{7}V\right)
+
5V_D
=
V.
\]
Thus,
\[
5V_D
=
V+\frac{8}{7}V
=
\frac{15}{7}V.
\]
Hence,
\[
V_D
=
\frac{3}{7}V.
\]
\[
\boxed{V_C=\frac{4V}{7},\qquad V_D=\frac{3V}{7}}
\]
Step 6: Calculate the total current supplied by terminal A.
Current from A to C:
\[
I_{AC}
=
\frac{V-V_C}{10}
=
\frac{V-\frac{4V}{7}}{10}
=
\frac{3V}{70}.
\]
Current from A to D:
\[
I_{AD}
=
\frac{V-V_D}{20}
=
\frac{V-\frac{3V}{7}}{20}
=
\frac{4V}{140}
=
\frac{V}{35}.
\]
Therefore,
\[
I
=
I_{AC}+I_{AD}.
\]
Substituting,
\[
I
=
\frac{3V}{70}
+
\frac{V}{35}.
\]
Taking the LCM,
\[
I
=
\frac{3V}{70}
+
\frac{2V}{70}
=
\frac{5V}{70}.
\]
Thus,
\[
I=\frac{V}{14}.
\]
Step 7: Determine the equivalent resistance.
Using Ohm's law,
\[
R_{AB}
=
\frac{V}{I}.
\]
Substituting
\[
I=\frac{V}{14},
\]
we obtain
\[
R_{AB}
=
\frac{V}{V/14}
=
14\,\Omega.
\]
\[
\boxed{R_{AB}=14\,\Omega}
\]
Final Answer:
The effective resistance between points A and B is
\[
\boxed{14\,\Omega}.
\]
Hence, the correct option is
\[
\boxed{\text{(A)}\ 14\,\Omega}.
\]
