Answer:

R= 200 ohm

I = 1 ampere

Solution-

[tex] \red{p = {i}^{2} r}[/tex]

[tex]p = 1 \times 1 \times 200[/tex]

[tex] \fbox{p = 200 \: watt}[/tex]

Potential diference-

[tex] \blue{v = ri} \: \: \: \: \: (ohms \: law)[/tex]

[tex]v = 1 \times 200 \\ \fbox{v = 200volt}[/tex]

Energy-

[tex] \green{e = p \times t }\\ = \frac{200}{1000} \times 5[/tex]

[tex] = \frac{2}{10} \times 5 \\ \fbox{= 1 \: kilo \: watt \: hour}[/tex]

Plz thanks

Verified answer

Power of the bulb,...

[tex]\begin{array}{l} P = I^{2}R \\ \Rightarrow (1)^{2} * 200\\\Rightarrow 200 \end{array}[/tex]

P = 200 W

Energy consumed by the bulb in 5h in burning = Power * Time

[tex]\begin{array}{l} \Rightarrow 200 * 5\\ \Rightarrow 1000 \end{array}[/tex]

Wh or 1k Wh

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