Question

Bulb B1 B2 and B3 have rated power 10W 10W and 15W respectively and rated voltage as 800V each. If the applied voltage is 200V then find power of the circuit?

Answer 1

Answer:

1 W

Explanation:

Given:

$B_1: 10W, 800V$     ;     $B_2: 10W, 800V$     ;     $B_3: 15W, 800V$

$V_{applied}=200V$

Power (P) =?

Step 1: Find $R_{effective}$:

$R = \frac{V^2}{P}$

$R_1 = \frac{800^2}{10}$               ;        $R_2 = \frac{800^2}{10}$         ;        $R_3 = \frac{800^2}{15}$

(Note: The question does not mention how the bulbs are arranged. I have answered the question using the image attached.)

$B_2$ and $B_3$ are connected in series i.e. their resistances are added:

$R_{2,3} = \frac{800^2}{10} + \frac{800^2}{15} = \frac{3(800)^2+2(800)^2}{30} = \frac{5(800)^2}{30} = \frac{800^2}{6}$

$B_1$ is connected in parallel i.e. their reciprocals are added:

$\frac{1}{R_{eff}} = \frac{6}{800^2} + \frac{10}{800^2} = \frac{16}{800^2}$

$R_{eff} = \frac{800^2}{16}$

Step 2: Find $P_{effective}$:

$P_{eff} = \frac{(V_{applied})^2}{R_{eff}}$

$P_{eff} = \frac{200\times200}{\frac{800^2}{16}} = \frac{200\times200\times16}{800^2} = 1$

∴ Effective power through the circuit is 1 W.

Any corrections or suggestions for improvement are welcome :)