A resistor of value R is connected across a voltage V. The minimum number of resistors of any value to be connected in parallel with it so that the current drawn from V is thrice the original value is

Concept:

When resistors are connected in parallel, the total (equivalent) resistance decreases, thereby increasing the current drawn from the voltage source.

Let the original resistor be of value R. The original current drawn from voltage V is:

\( I = \frac{V}{R} \)

We want the total current to be thrice the original:

\( I_{new} = 3 \cdot \frac{V}{R} = \frac{V}{R_{eq}} \Rightarrow R_{eq} = \frac{R}{3} \)

Calculation:

Now, we need to find the minimum number of resistors (of any value) that can be added in parallel with R to make the equivalent resistance = R/3.

Let one resistor of value X be added in parallel with R:

\( \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{X} \)

We want \( R_{eq} = \frac{R}{3} \Rightarrow \frac{1}{R_{eq}} = \frac{3}{R}\)

\( \frac{3}{R} = \frac{1}{R} + \frac{1}{X} \Rightarrow \frac{2}{R} = \frac{1}{X} \Rightarrow X = \frac{R}{2} \)

So, adding one resistor of value R/2 in parallel will reduce the total resistance to R/3 and increase the current to 3I.

Final Answer:

Option 1) One