Question

The voltage drop across \( 15 \, \Omega \) resistance in the given figure will be \(\dots\dots\) V. 

Hint:

When parallel branches have identical total resistance, the main current divides equally among them, significantly simplifying the node voltage calculations.

Answer 1

Correct Answer: 6

Step 1: Understanding the Concept:
In steady DC analysis, we calculate the equivalent resistance of the circuit to find the main current. Then, using current division rules, we find the current in specific branches to determine the voltage drop across individual resistors using Ohm's Law (\( V = IR \)).
Step 2: Key Formula or Approach:
1. Equivalent resistance of parallel resistors: \( R_p = \frac{R_1 R_2}{R_1 + R_2} \)
2. Total current: \( I = \frac{V}{R_{eq} + r} \)
Step 3: Detailed Explanation:
Let's simplify the parallel and series combinations in the upper and middle branches:
1. Upper Branch (\( R_{top} \)):
- First parallel part: \( 4 \, \Omega || 4 \, \Omega = 2 \, \Omega \).
- Second part: \( 2 \, \Omega \) in series.
- Third parallel part: \( 15 \, \Omega || 10 \, \Omega = \frac{15 \times 10}{15 + 10} = \frac{150}{25} = 6 \, \Omega \).
- Total resistance of upper branch: \( R_{top} = 2 + 2 + 6 = 10 \, \Omega \).
2. Middle Branch (\( R_{mid} \)):
- First parallel part: \( 8 \, \Omega || 8 \, \Omega = 4 \, \Omega \).
- Second parallel part: \( 12 \, \Omega || 12 \, \Omega = 6 \, \Omega \).
- Total resistance of middle branch: \( R_{mid} = 4 + 6 = 10 \, \Omega \).
3. Total Resistance and Current:
- These two branches are in parallel: \( R_{eq\_parallel} = \frac{10 \times 10}{10 + 10} = 5 \, \Omega \).
- Adding the internal/series resistor (\( 1 \, \Omega \)): \( R_{total} = 5 + 1 = 6 \, \Omega \).
- Total current from battery: \( I_{total} = \frac{12 \, V}{6 \, \Omega} = 2 \, A \).
4. Current Division:
- Since \( R_{top} = R_{mid} = 10 \, \Omega \), the current splits equally: \( I_{top} = 1 \, A \) and \( I_{mid} = 1 \, A \).
- The voltage drop across the \( 15 \, \Omega \) resistor is the voltage drop across the entire \( 15 || 10 \) parallel combo:
\[ V_{15 \Omega} = I_{top} \times R_{p2} = 1 \, A \times 6 \, \Omega = 6 \, V \]
Step 4: Final Answer:
The voltage drop across the \( 15 \, \Omega \) resistance is \( 6 \, V \).