Question

Two electric bulbs have resistances R and 3R respectively. They are connected in parallel to a power source of voltage V. If the two resistors are to be replaced by a single resistor. What will be the value of this resistor? *
1 point
0.75 R
1.33 R
4 R
Insufficient information

Answer 1

Answer:

Resistors = R and 3R

It is given that the bulbs with resistance R and 3R are connected in parallel connection across a voltage difference 'V'.

Now if these resistances were to be replaced by a bulb with single resistance, what would be it's value is the question.

Considering the resistors alone in a parallel combination, we get the effective resistance to be:

$\implies \dfrac{1}{R_{eff}} = \dfrac{1}{3} + \dfrac{1}{1}\\\\\\ \implies R_{eff} = \dfrac{3 \times 1}{3 + 1}\\\\\\ \implies \boxed{\bf{R_{eff} = \dfrac{3}{4} = 0.75 \Omega}}$

Hence the effective resistance is 0.75 Ω.

Hence the value of the single resistance must be 0.75 Ω.

Answer 2

Given :-

Two electric bulbs have resistances R and 3R respectively. They are connected in parallel to a power source of voltage V. If the two resistors are to be replaced by a single resistor

To Find :-

What will be the value of this resistor

Solution :-

We know that

${\boxed{\frak{\underline{\ast \dfrac{1}{R_p}=\dfrac{1}{R1}+\dfrac{1}{R2}\ast}}}}$

We have

R1 = 1R

R2 = 3R

$\sf:\implies \dfrac{1}{R_p}=\dfrac{1}{1R}+\dfrac{1}{3R}$

$\sf:\implies \dfrac{1}{R_p}=\dfrac{3 + 1}{3R}$

$\sf:\implies \dfrac{1}{R_p}=\dfrac{4}{3R}$

On cross multiplication

$\sf:\implies 3R(1)=4(R_p)$

$\sf:\implies 3R= 4R_p$

$\sf:\implies\dfrac{3R}{4}=R_p$

$\sf:\implies 0.75R=R_p$

Hence

The value of reistor is 0.75 R Option A is correct