Question

(A) Define electric power. Express it in terms of potential difference V and resistance R.
(B) An electrical fuse is rated at 2A. What is meant by this statement ?
(C) An electric iron of 1 kW is operated at 220V. Find which of the fuses that respectively rated at 1A, 3A and 5A can be used in it.

Answer 1

Hint :
For solving part A, first write the definition of electric power and derive the formula of electric power using ohm’s law.
$V = IR$
Where
V $ = $ Potential difference
I $ = $ Current flowing in circuit
R $ = $ Resistance of circuit
For part B, using heating effect at current for part C, first calculate the current drawn by using following expression
$I = \dfrac{P}{V}$
Where
P $ = $ Power
V $ = $ Voltage $($Potential difference$)$ of source
After 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 volt
So, current drawn $I = \dfrac{P}{V}$
$I = \dfrac{{1000}}{{220}} = \dfrac{{50}}{{11}}$
$I = 4.54$ Amp
Hence, 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 are
1. $P = VI$ $($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 $P = \dfrac{{{V^2}}}{R}$