Question

2. 110V and 100W in an electric bulb
Has been recorded.
A. What is the power consumption of this device per second?
b. What is the resistance of this bulb?​

Answer 1

Answer:

A.w=pt=100 w.1 s=100 joule

B.here,

power,p=100 w

voltage,v=110 v

Resistance, R=?

we know that,

p=v^2/R

or,R=v^2/p

or,R=110^2/100=121 oh'm

Answer 2

Explanation:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\power( P) = 100 \: W \\ potential \: difference(V) = 110V \\ \blue{\mathfrak{\underline{\large{Finding}}}:} \\ (A): \: P= \frac{energy(joules)}{time(second)} = \frac{E(J)}{t(s)} = 100J {s}^{ - 1} \\(B): \: resistance(R) = ? \\ P= \frac{ {V}^{2} }{R} \\ R = \frac{ {V}^{2} }{P} \\ = \frac{ {(110)}^{2} }{100} \\ \boxed{R = 121 \: Ω} \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$