Three electric bulbs of 200,W, 200,W and 400, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the two bulbs of $200\,W$ be $A$ and $B$,and the bulb of $400\,W$ be $C$.Bulbs $A$ and $B$ are connected in parallel. The effective power $P^{

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(C) Let the two bulbs of $200\,W$ be $A$ and $B$,and the bulb of $400\,W$ be $C$.Bulbs $A$ and $B$ are connected in parallel. The effective power $P^{\prime}$ of this parallel combination is given by the sum of their individual powers:$P^{\prime} = P_{A} + P_{B} = 200\,W + 200\,W = 400\,W$.Now,this combination $P^{\prime}$ is in series with bulb $C$ $(400\,W)$. The resultant power $P_{R}$ of two components in series is given by the formula $\frac{1}{P_{R}} = \frac{1}{P^{\prime}} + \frac{1}{P_{C}}$,which simplifies to $P_{R} = \frac{P^{\prime} 	imes P_{C}}{P^{\prime} + P_{C}}$.Substituting the values:$P_{R} = \frac{400\,W 	imes 400\,W}{400\,W + 400\,W} = \frac{160000\,W^2}{800\,W} = 200\,W$.

Three electric bulbs of $200\,W, 200\,W$ and $400\,W$ are shown in the figure. The resultant power of the combination is ................ $W$.

Similar Questions

The equivalent resistance between points $A$ and $B$ in the given network is ............ $\Omega$.

The figure shows three resistor configurations $R_1, R_2$, and $R_3$ connected to a $3 \text{ V}$ battery. If the power dissipated by the configurations $R_1, R_2$, and $R_3$ is $P_1, P_2$, and $P_3$ respectively, then:

In the given network,the equivalent resistance between $A$ and $B$ is .............. $\Omega$.

$A$ piece of wire is cut into four equal parts and the pieces are bundled together side by side to form a thicker wire. Compared with that of the original wire,the resistance of the bundle is ..........

Equivalent resistance of the following network is . . . . . . $\Omega$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module