Question

An electric bulb of resistance 200ohm draws a current of 1 Ampere. Calculate

(i) the power of the bulb

(ii) the potential difference at its ends and

(iii) the energy in kWh consumed burning it for 5h.​

Answer 1

Step-by-step explanation:

given that,

R=200ohm

i=1A

power= i^2×R

so power= 1×200 watt

:: P=200 watt

potential difference = i×R

V= 200 volts

Energy= power×time

E= (200×5)/1000 kwh

ENERGY = 1 KWh

Answer 2

Answer: 1. 200 W

              2. 200 V

              3. 1 kWh

Formula used:

$Power(P)=Current(I)^{2} *Resistance(R)$

$Potential(V)=Current(I)*Resistance(R)$

$Energy(E)=Power(P)*Time(t)$

Explanation:

Given,

Resistance (R)=200Ω

Current (I)=1 A

Time (t) = 5 hrs.

Power : It is the amount of energy transferred per unit time.

1.  P=1²×200

   P=200 W

Potential difference: The voltage difference between any two points in a circuit is known as Potential Difference .

2. V=1×200

   V=200 V

Energy : It is the capacity for doing work.

3. E=200×5                      (1 kWh = 1000 W so, 200 W = 0.2 kWh)

   E=0.2×5

  E=1 kWh