Question

In the network shown below, if potential across XY is 4 V, then the input potential across AB is

• A. 16 V
• B. 20 V
• C. 8 V
• D. 12 V
• E. 24 V

Hint:

In complex circuit analysis, always start from the part of the circuit where the most information is provided (like the voltage at XY) and work your way backwards toward the input source.

Answer 1

The Correct Option is B

Concept: Solving complex resistor networks involves identifying series and parallel combinations and applying Ohm's Law ($V = IR$) step-by-step.
• Parallel Resistance: For identical resistors $R$ in parallel, $R_{eq} = R/n$.
• Current Distribution: In parallel branches, current divides inversely to resistance. In series, current remains constant.
• Potential Drop: Total potential is the sum of drops across series components ($V = V_1 + V_2 + \dots$).

Step 1: Analyze the rightmost section near XY.
The potential across the $4\ \Omega$ resistor (XY) is given as $4\text{ V}$. Using Ohm's Law ($I = V/R$), the current through this branch is: \[ I_{XY} = \frac{4\text{ V}}{4\ \Omega} = 1\text{ A} \] Since there are two $4\ \Omega$ resistors in parallel at the end, each carries $1\text{ A}$, making the total current entering that junction $2\text{ A}$.

Step 2: Simplify the central parallel network.
The two $8\ \Omega$ resistors are in parallel. Their equivalent resistance is: \[ R_p = \frac{8 \times 8}{8 + 8} = 4\ \Omega \] The current of $2\text{ A}$ from the previous step passes through this $4\ \Omega$ equivalent resistance, creating a potential drop of: \[ V = 2\text{ A} \times 4\ \Omega = 8\text{ V} \] This $8\text{ V}$ is also across the vertical $4\ \Omega$ resistor on the left of that section. The current through it is $I = 8\text{ V} / 4\ \Omega = 2\text{ A}$.

Step 3: Calculate total input potential AB.
The total current coming from point A is the sum of the branch currents: $2\text{ A} + 2\text{ A} = 4\text{ A}$. Potential drop across input $2\ \Omega$ resistors at both top and bottom: $V = 4\text{ A} \times (2\ \Omega + 2\ \Omega) = 16\text{ V}$. The potential across the central parallel bridge was determined to be $4\text{ V}$ based on the XY output. \[ V_{AB} = 16\text{ V} + 4\text{ V} = 20\text{ V} \]