If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Given: An electric bulb is rated 220V and 100W Find out: Power consumed when it is operated on 110 V Explanation: In physics, an electric power measure of the rate of electrical energy transfer by an electric circuit per unit time. Denoted by P and measured using the SI unit of power is the watt or one joule per second. Electric power is commonly supplied by sources such as electric batteries and produced by electric generators. The formula for electric power is given by where, P = Power V = Potential difference in the circuit I= Electric current SolutionAs per the equation P = V2/R we have If V = 220 V 100 W = 2202/R R = 2202/100 Ω = 484 Ω. This is the resistance of the bulb. When V = 110 V, power consumed = V2/R = 1102/484 = 25 W. So, 25 W power is consumed when it is operated on 110 V. Articles to Explore:
(i) Given: Power of the bulb, $P_B=40W$ Voltage, $V=220V$ To find: Current drawn by the bulb, $I$. Solution: We know that electric power is given as- $P=V\times I$ Therefore, $I=\frac {P}{V}$ Substituting the required value, we get- $I=\frac {40}{220}$ $I=0.18A$ Thus, the current drawn by the bulb is 0.18A. We know that resistance is given as- $R=\frac {V^2}{P}$ Putting the required value we get- $R=\frac {220^2}{40}$ $R=\frac {220\times 220}{40}$ $R=1210\Omega$ Thus, the resistance of the 40 W bulb is 1210 ohm. Now, the given bulb is replaced by a bulb of rating 25 W; 220 V. Given: Power of the bulb, $P_B=25W$ Voltage, $V=220V$ To find: Current drawn by the bulb, $I$, and resistance, $R$. Solution: We know that electric power is given as- $P=V\times I$ Therefore, $I=\frac {P}{V}$ Substituting the required value, we get- $I=\frac {25}{220}$ $I=0.113A$ Thus, the current drawn by the bulb is 0.113A. We know that resistance is given as- $R=\frac {V^2}{P}$ Putting the required value we get- $R=\frac {220^2}{25}$ $R=\frac {220\times 220}{25}$ $R=1936\Omega$ Thus, the resistance of the 25 W bulb is 1936 ohm. Yes, there is a change in current and resistance. $\Rightarrow change\ in\ current=0.18-0.1136=0.0664\ A$ $\Rightarrow change\ in\ resistance=1936-1210=726\Omega$ Hence, from the above justification, we can see that current decreases and resistance increases when we use a 25Watt bulb in place of a 40Watt. |
An electric bulb is rated 40 W, 200 v If it is operated at 120 V then power consumed by it is
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