A 100 W bulb is rated at 200 V and a 200 W | vedclass

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

Question

(B) The resistance $R$ of a bulb is given by the formula $R = \frac{V^2}{P}$.For the first bulb $(P_1 = 100 \ W, V_1 = 200 \ V)$: $R_1 = \frac{200^2}{

Vedclass · Accepted Answer

The ratio of their maximum current ratings is $1 : 4$.

Vedclass · Answer

(B) The resistance $R$ of a bulb is given by the formula $R = \frac{V^2}{P}$.For the first bulb $(P_1 = 100 \ W, V_1 = 200 \ V)$: $R_1 = \frac{200^2}{100} = \frac{40000}{100} = 400 \ \Omega$.For the second bulb $(P_2 = 200 \ W, V_2 = 100 \ V)$: $R_2 = \frac{100^2}{200} = \frac{10000}{200} = 50 \ \Omega$.The ratio of resistances is $\frac{R_1}{R_2} = \frac{400}{50} = 8 : 1$.The maximum current rating $i$ is given by $i = \frac{P}{V}$.For the first bulb: $i_1 = \frac{100}{200} = 0.5 \ A$.For the second bulb: $i_2 = \frac{200}{100} = 2 \ A$.The ratio of maximum current ratings is $\frac{i_1}{i_2} = \frac{0.5}{2} = \frac{1}{4}$.

$A$ $100 \ W$ bulb is rated at $200 \ V$ and a $200 \ W$ bulb is rated at $100 \ V$. Then:

Similar Questions

$A$ $25\, W$, $220\, V$ bulb and a $100\, W$, $220\, V$ bulb are connected in parallel across a $440\, V$ line. Which of the following will happen?

Above neutral temperature,thermo $e.m.f.$ in a thermocouple

The thermo emf of a hypothetical thermocouple varies with the temperature $\theta$ of the hot junction as $E = a\theta + b\theta^2$ in volts,where the ratio $a/b$ is $700^{\circ}C$. If the cold junction is kept at $0^{\circ}C$,then the neutral temperature is:

The maximum power delivered to resistance $R$ is ............... $W$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module