Power of Electrical Circuit: Practice Problem & Solution
A filament bulb ($500\text{ W}$, $100\text{ V}$) is to be used in a $230\text{ V}$ main supply. When a resistance $R$ is connected in series, it works perfectly and the bulb consumes $500\text{ W}$. The value of $R$ is: (2016 - II)
Solution Explained:
To solve this problem, we apply the core principles of Power of Electrical Circuit. Understanding the underlying formula is key to arriving at the correct answer below:
Rated current for the bulb is $I = \frac{P}{V_b} = \frac{500}{100} = 5\text{ A}$.
To operate on a $230\text{ V}$ supply, potential drop across $R$ must be $V_R = 230 - 100 = 130\text{ V}$.
Therefore, $R = \frac{V_R}{I} = \frac{130}{5} = 26\text{ }\Omega$.
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