Question

an electric lamp is marked 25w , 220v it is used for 10 h daily calculate. (a) resistance while glowing (b) energy consumed in kwh per day .​

Answer 1

Given:-

P = 25w

1000 w = 1kw

25 w = 1/1000 × 25

= 0.025 kw

V = 220 v

T = 10h

Answer:-

$R = 1936\Omega$

$E = 0.25 kwh$

To find:-

a) Resistance while glowing.

b) Energy consumed in kwh / day.

Solution:-

Resistance of lamp is given by:-

$\boxed{\sf{P = \dfrac{V^2}{R}}}$

Now, put the given value,

$\implies$ $25 = \dfrac{(220)^2}{R}$

$\implies$ $25 = \dfrac{48,400}{R}$

$\implies$ $25R = 48,400$

$\implies$ $R = \dfrac {48,400}{25}$

$R=1,936 \Omega$

Energy consumed is given by :-

$\boxed{\sf{E = P. T}}$

$\implies$ $E = 0.025 × 10$

$E = 0.25 kwh$

hence, the resistance of lamp is 1936 ohm and energy consumed is 0.25 kwh.

Answer 2

(a) Resistance of lamp = 1936Ω or 1.936 k Ω

(b) Energy consumed per day = 0.25 kw h

Explanation:

Given, Power = 25 W

Voltage = 220 V

Time for which it is used in a day = 10 h

(a) We know that Power = voltage * current

=> P = V I

Also V= IR

=> P = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$

=> R = $\frac{220^{2} }{25}$

=> R = 1936 Ω

or R = 1.936 k Ω

(b) Power = $\frac{Energy}{time}$

=> Energy = Power * time

=> Energy = 25 * 10

=>               = 250 w h

=> Energy= $\frac{250}{1000}$  kW h

=> Energy = 0.25 k w h

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