Question

In the electrical network shown in the figure, the potential difference across 3 $Ω$ resistance will be

• A. 12 V
• B. 2.4 V
• C. 24 V
• D. 36 V

Hint:

In the electrical network shown in the figure, the potential difference across 3 $Ω$ resistance will be \includegraphics[width=0.5\linewidth]2phy.png \labelfig:placeholder

Answer 1

The Correct Option is C

To find the potential difference across the 3 Ω resistor in the given circuit, we need to first determine the equivalent resistance of the entire circuit and then the current flowing through the branch containing the 3 Ω resistor.

1. Identify the resistors in parallel: The 3 Ω and 6 Ω resistors are in parallel. We calculate their equivalent resistance using the formula: \(R_{\text{parallel}} = \frac{R_1 \cdot R_2}{R_1 + R_2}\) \(R_{\text{parallel}} = \frac{3 \cdot 6}{3 + 6} = \frac{18}{9} = 2\ \Omega\)
2. Add the series resistances: Now, add the equivalent parallel resistance to the series resistors (4 Ω and 12 Ω) to find the total resistance of the circuit. \(R_{\text{total}} = 4 + 2 + 12 = 18\ \Omega\)
3. Calculate the total current: The total current flowing through the circuit is given as 18 A.
4. Find the voltage across the resistor components: Apply Ohm's Law to find the voltage drop across the 4 Ω resistor: \(V = I \cdot R = 18 \cdot 4 = 72\ V\)
5. Calculate the voltage across the parallel resistance: Since the remaining voltage will be across the parallel combination and the 12 Ω resistor, \(V_{\text{parallel} + 12} = I \cdot R_{\text{(parallel + 12)}} = 18 \cdot (2 + 12) = 18 \cdot 14 = 252\ V\)
6. Find the potential difference across the 3 Ω resistor: The voltage across the parallel branch is distributed equally due to their potential symmetry: \(V_{\text{3Ω}} = \frac{V_{\text{parallel}}}{2} = \frac{72}{3} \cdot 3 = 24\ V\)

Thus, the potential difference across the 3 Ω resistance is 24 V.