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

Given:
Resistance of each bulb, \( R = 8 \, \Omega \)
Battery voltage, \( V = 4 \, \text{V} \)
Total current required, \( I = 2 \, \text{A} \)

Step 1: Find the Equivalent Resistance Required
Using Ohm’s law: \[ V = IR \] \[ R_{\text{eq}} = \frac{V}{I} \] \[ R_{\text{eq}} = \frac{4}{2} \] \[ R_{\text{eq}} = 2 \, \Omega \] So, the equivalent resistance of the parallel combination must be \( 2 \, \Omega \).

Step 2: Use Formula for Parallel Combination
For \( n \) identical resistors connected in parallel: \[ R_{\text{eq}} = \frac{R}{n} \] Substitute values: \[ 2 = \frac{8}{n} \]

Step 3: Solve for \( n \)
\[ n = \frac{8}{2} \] \[ n = 4 \]

Final Answer: \[ \boxed{\text{Number of bulbs required} = 4} \]
Therefore, four bulbs of 8 Ω each must be connected in parallel to draw 2 A current from a 4 V battery.