A 100, W bulb B_1 and two 60, W bulbs B_2 and | vedclass

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

Question

(D) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For bulb $B_1$ $(100\, W)$: $R_1 = \frac{V^2}{100}$.For bulbs $B_2$ and $B_3$ $(60\, W)$:

Vedclass · Accepted Answer

$P_1 < P_2 < P_3$

Vedclass · Answer

(D) The resistance of a bulb is given by $R = \frac{V^2}{P}$.For bulb $B_1$ $(100\, W)$: $R_1 = \frac{V^2}{100}$.For bulbs $B_2$ and $B_3$ $(60\, W)$: $R_2 = R_3 = \frac{V^2}{60}$.From the circuit diagram,bulbs $B_1$ and $B_2$ are in series,and this combination is in parallel with bulb $B_3$ across the $220\, V$ source.Bulb $B_3$ is connected directly across the $220\, V$ source,so its power $P_3 = 60\, W$.For the series combination of $B_1$ and $B_2$,the current $I = \frac{V}{R_1 + R_2} = \frac{V}{\frac{V^2}{100} + \frac{V^2}{60}} = \frac{V}{V^2(\frac{3+5}{300})} = \frac{300}{8V} = \frac{37.5}{V}$.The power consumed by $B_1$ is $P_1 = I^2 R_1 = (\frac{37.5}{V})^2 \cdot \frac{V^2}{100} = \frac{1406.25}{100} = 14.06\, W$.The power consumed by $B_2$ is $P_2 = I^2 R_2 = (\frac{37.5}{V})^2 \cdot \frac{V^2}{60} = \frac{1406.25}{60} = 23.44\, W$.Comparing the values: $P_1 = 14.06\, W$,$P_2 = 23.44\, W$,and $P_3 = 60\, W$.Therefore,$P_1 < P_2 < P_3$.

$A$ $100\, W$ bulb $B_1$ and two $60\, W$ bulbs $B_2$ and $B_3$ are connected to a $220\, V$ source,as shown in the figure. If $P_1, P_2$,and $P_3$ are the output powers of the bulbs $B_1, B_2$,and $B_3$ respectively,then:

Similar Questions

Lamps used for household lighting are connected in

The power dissipated in the circuit shown in the figure is $30 \, W$. The value of $R$ is ............. $\Omega$.

The equivalent resistance of the circuit shown below between points $a$ and $b$ is $..........\Omega$

When two resistances are connected in parallel,the equivalent resistance is $\frac{6}{5} \, \Omega$. When one of the resistances is removed,the effective resistance is $2 \, \Omega$. The resistance of the wire removed will be ................ $\Omega$.

For three resistors of different values connected in series,obtain the equation for equivalent resistance. From this,write the equation for $n$ resistors connected in series.

Vedclass Test Series

Exam Paper Generator

Online Exam Module