MCQ 11 Mark
An electrical appliance has a rating 100 W, 120V. The resistance of element of appliance when in use:
- A1.2 Ω
- ✓144 Ω
- C120 Ω
- D100 Ω
Answer
View full question & answer→Correct option: B.
144 Ω
$144 \Omega$
Given, Power $(P)=100 W$
Potential difference , V = 120 volt
$\therefore$ Resistance, $R =\frac{ V ^2}{ P }=\frac{(120)^2}{100}=144 \Omega$
Given, Power $(P)=100 W$
Potential difference , V = 120 volt
$\therefore$ Resistance, $R =\frac{ V ^2}{ P }=\frac{(120)^2}{100}=144 \Omega$










