A 25, W,220, V bulb and a 100, W,220, V bulb | vedclass

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

Question

(A) The resistance $R$ of a bulb is given by $R = V^2 / P$,where $V$ is the rated voltage and $P$ is the rated power.For the $25\, W$ bulb,$R_1 = (220

Vedclass · Accepted Answer

$25\, W$ bulb

Vedclass · Answer

(A) The resistance $R$ of a bulb is given by $R = V^2 / P$,where $V$ is the rated voltage and $P$ is the rated power.For the $25\, W$ bulb,$R_1 = (220)^2 / 25 = 1936\, \Omega$.For the $100\, W$ bulb,$R_2 = (220)^2 / 100 = 484\, \Omega$.When connected in series,the current $I$ flowing through both bulbs is the same.The brightness of a bulb is proportional to the power dissipated,given by $P_{dissipated} = I^2 R$.Since $R_1 > R_2$,the power dissipated in the $25\, W$ bulb is greater than that in the $100\, W$ bulb.Therefore,the $25\, W$ bulb will glow more brightly.

$A$ $25\, W$,$220\, V$ bulb and a $100\, W$,$220\, V$ bulb are connected in series across a $220\, V$ line. Which electric bulb will glow more brightly?

Similar Questions

Two sources of equal $emf$ $E$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

In the given circuit, what is the reading of the voltmeter in $V$?

The readings of the voltmeter and ammeter in the circuit shown in the diagram are respectively

Each element in the finite chain of resistors shown in the figure is $1\,\Omega $. $A$ current of $1\, A$ flows through the final element. Then what is the potential difference $V$ across input terminals of the chain (in $,V$)?

In the given electric circuit, the potential difference across $PQ$ is approximately ...... volts.

Vedclass Test Series

Exam Paper Generator

Online Exam Module