If R-1 and R-2 are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage- thenR-1 - 2R-2R-2 - 2R-1R-2 - 4R-1R-1 - 4R-2

Question

Toppr Admin · Accepted Answer

Given that-Power of bulbs-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- P1-200-xA0-WP2-100-xA0-WThe power P-VI- and-xA0- -xA0- I-V-R-xA0-xA0-On substituting we get-xA0-xA0-xA0- -xA0-P-V2R-Here- the power P1 used in case of R1 is P1-V2R1-200W-1-

If $R_{1}$ and $R_{2}$ are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, then
$R_{1} = 2 R_{2}$
$R_{2} = 2 R_{1}$
$R_{2} = 4 R_{1}$
$R_{1} = 4 R_{2}$

Given that,
Power of bulbs $P_{1} = 200 W P_{2} = 100 W$

The power $P = V I$ , and $I = V / R$

On substituting we get, $P = \frac{V^{2}}{R}$ .

Here, the power $P_{1}$ used in case of $R_{1}$ is $P_{1} = \frac{V^{2}}{R_{1}} = 200 W$ ..........(1)

If R1 and R2 are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, thenR1=2R2R2=2R1R2=4R1R1=4R2

Given that,Power of bulbs P1=200 WP2=100 WThe power P=VI, and I=V/R On substituting we get, P=V2R.Here, the power P1 used in case of R1 is P1=V2R1=200W..........(1)

If $R_{1}$ and $R_{2}$ are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, then
$R_{1} = 2 R_{2}$
$R_{2} = 2 R_{1}$
$R_{2} = 4 R_{1}$
$R_{1} = 4 R_{2}$

Given that,
Power of bulbs $P_{1} = 200 W P_{2} = 100 W$

The power $P = V I$ , and $I = V / R$

On substituting we get, $P = \frac{V^{2}}{R}$ .

Here, the power $P_{1}$ used in case of $R_{1}$ is $P_{1} = \frac{V^{2}}{R_{1}} = 200 W$ ..........(1)