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(a) Electric power: It is the rate of doing work by an energy source or the rate at which the electrical energy is dissipated or consumed per unit time in the electric circuit is called electric power. (b) It means, the maximum current will flow through it is only 2 A. Fuse wire will melt if the current exceeds 2 A value through it. Please log in or register to add a comment.
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. Power in electronics is the rate of doing work. Resistance is the opposition offered against the flow of electrons. The relation between power and resistance is proportional. In physics, power and resistance can be related using two formulas. The relation between Power and Resistance can be mathematically expressed in two ways, as follows- Where, \(\begin{array}{l}P=\frac{V^{2}}{R}\end{array} \) Where,
Power and Resistance RelationBoth the formula mentioned above gives the relation between power and resistance. Consider P = I2R the electric power is directly proportional to resistance keeping I constant. Which implies-
Consider \(\begin{array}{l}P=\frac{V^{2}}{R}\end{array} \) here, Power P is inversely proportional to Resistance R. which implies that-For any constant Potential difference
Power and Resistance FormulaDeriving Power and Resistance formula will concrete the understanding of the concept. Electric power(P) is the measure of the electric current I with Q coulombs of charge passing through a potential difference of V volts in time t seconds. Mathematically given by- \(\begin{array}{l}P=\frac{VQ}{t}=IV\end{array} \) ——-(1)From Ohm’s law, we know that V=IR Rearranging it for I we get, \(\begin{array}{l}I=\frac{V}{R}\end{array} \) Thus, substituting it in equation (1) we get- \(\begin{array}{l}P=\frac{V^{2}}{R}\end{array} \) Hope you have understood the relation between Power and Resistance, formulas along with the units of electricity and derivation. Also, power and resistance conversion in physics and electronics. Physics Related Topics: Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S-The Learning App” for loads of interactive, engaging physics-related videos and an unlimited academic assist. What is Ohm’s Law?
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out of 0 arewrong out of 0 are correct out of 0 are Unattempted View Quiz Answers and AnalysisAnswer Hint :For solving part A, first write the definition of electric power and derive the formula of electric power using ohm’s law.$V = IR$WhereV $ = $ Potential difference I $ = $ Current flowing in circuitR $ = $ Resistance of circuitFor part B, using heating effect at current for part C, first calculate the current drawn by using following expression$I = \dfrac{P}{V}$WhereP $ = $ PowerV $ = $ Voltage $($Potential difference$)$ of sourceAfter comparing this to given values of current, we will get a desired solution. Complete step by step solution : (A) Electric power – Electric power is the rate per unit time at which electrical energy is transferred by an electric circuit.The SI unit of power is watt. Power is given by the product of potential difference V and current I i.e., $P = VI$Using ohm’s law $V = IR$$I = \dfrac{V}{R}$$\because P = V\left( {\dfrac{V}{R}} \right)$$P = \dfrac{{{V^2}}}{R}$Above expression shows the electric power in terms of potential difference V and resistance R.(B) In an electrical fuse is rated at 2 Amp. i.e., the maximum current will flow through it is only 2 Ampere. If current exceeds 2Amp then fuse wire will melt(C) Given that power P $ = $ 1 kW$P = 1000W \\ $And voltage V $ = $ 220 voltSo, current drawn $I = \dfrac{P}{V}$$I = \dfrac{{1000}}{{220}} = \dfrac{{50}}{{11}}$$I = 4.54$ AmpHence, to run an electric iron of 1 kW, rated fuse of 5 Amp should be used. Note : - Here, one thing should be noted that we have 2 formulas of power which are1. $P = VI$ $($In terms of V and I$)$2. $P = \dfrac{{{V^2}}}{R}$ $($In terms of V and R$)$- Students may get confused between these 2 formulas when and where they have to use them.- So, when electrical instruments are connected in series then current flowing in each instrument is the same. So, we should use $P = VI$- And then electric instruments are connected in parallel then voltage drop is the same. So, use $P = \dfrac{{{V^2}}}{R}$ |
Write electric power in terms of resistance and potential difference
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