Question

How many bulbs of resistance \( 8 \ \Omega \) each should be connected in parallel combination to draw a current of \( 2 \text{ A} \) from a battery of \( 4 \text{ V} \) ?

Hint:

In parallel, adding more resistors \textbf{decreases} the equivalent resistance and \textbf{increases} the total current drawn.

Answer 1

Step 1: Understanding the Concept:
Find the total required resistance of the circuit and then use the formula for parallel resistance to find the number of bulbs.
Step 2: Key Formula or Approach:
Ohm's Law: \( V = IR \implies R_{eq} = \frac{V}{I} \)
Parallel Resistance: \( \frac{1}{R_{eq}} = \frac{n}{r} \) (where \( n \) is the number of bulbs and \( r \) is the resistance of one bulb).
Step 3: Detailed Explanation:
1. Calculate Required Equivalent Resistance (\( R_{eq} \)):
\[ R_{eq} = \frac{V}{I} = \frac{4 \text{ V}}{2 \text{ A}} = 2 \ \Omega \]
2. Calculate number of bulbs (\( n \)):
Given individual resistance \( r = 8 \ \Omega \).
For \( n \) identical resistors in parallel:
\[ R_{eq} = \frac{r}{n} \]
\[ 2 = \frac{8}{n} \]
\[ n = \frac{8}{2} = 4 \]
Step 4: Final Answer:
4 bulbs should be connected in parallel.