Question

Two resistors of equal value are connected in parallel with each other. The power dissipated by the two resistors will be equal to the ? .

Answer 1

Answer:

Sum of the power dissipated by each resistor connected in parallel.

Solution:

As per the question:

Let the equal value resistor be of R' $\Omega$ each.

In parallel, the equivalent resistance is given by:

$\frac{1}{R_{e}} = \frac{1}{R'} + \frac{1}{R'}$

$R_{e} = \frac{R'^{2}}{2R'} = \frac{R}{2}$

The power dissipated in parallel is given by:

$P_{d} = \frac{V^{2}}{R_{e}}$

$P_{d} = \frac{V^{2}}{\frac{R}{2}}$

$P_{d} = \frac{2V^{2}}{R}$                                                 (1)

Also, it can be seen that the overall power dissipated in the resistors is equal to the sum of the power dissipated in each resistor.

Now,

Power dissipated in one resistor, P = $\frac{V^{2}}{R}$

Sum of the power dissipated in each resistor, P = $\frac{V^{2}}{R} + \frac{V^{2}}{R} = \frac{2V^{2}}{R}$         (2)

Thus from eqn (1) and (2), we can say that the sum of the power in the two resistors connected in parallel is equal.

Answer 2

Answer:

Explanation:

Let the two equal resistance be R. the voltage of the source is V.

When the tow resistance are connected in paralel, then the equivalent resistance is R/2 .

Power across this parallel combination

P = 2V²/R

Now when they are connected across the voltage source individually,

P1 = V²/R

P2 = V²/R

So, P = P1 + P2

Thus, the total power is the sum of the individual powers.