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Q.
An electric bulb is marked $100 W , 230 V$. If the supply voltage drops to $115 V$, what is the heat and light energy produced by the bulb in $20 min$ ?
Current Electricity
Solution:
$R=\frac{V^{2}}{P}=\frac{(230)^{2}}{100}=529 \Omega$
When the voltage drops to $115 V$, heat and light energy produced by the bulb in $20 min$ is given by
$W=\frac{V^{2}}{R} t=\frac{(115)^{2}}{529} \times 20 \times 60=30000 J$