Respuesta :
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' [tex]\Omega[/tex] each.
In parallel, the equivalent resistance is given by:
[tex]\frac{1}{R_{e}} = \frac{1}{R'} + \frac{1}{R'}[/tex]
[tex]R_{e} = \frac{R'^{2}}{2R'} = \frac{R}{2}[/tex]
The power dissipated in parallel is given by:
[tex]P_{d} = \frac{V^{2}}{R_{e}}[/tex]
[tex]P_{d} = \frac{V^{2}}{\frac{R}{2}}[/tex]
[tex]P_{d} = \frac{2V^{2}}{R}[/tex] (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 = [tex]\frac{V^{2}}{R}[/tex]
Sum of the power dissipated in each resistor, P = [tex]\frac{V^{2}}{R} + \frac{V^{2}}{R} = \frac{2V^{2}}{R}[/tex] (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:
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.
