Question #0ad4c

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Question #0ad4c

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Answer 1 · 2015-06-18T13:14:32

0.5681 amps

Explanation:

Arunraju-n

We must find out electrical resistance if lamp load and their ratings are given.

current in circuit= current in lamp1 + current in lamp2

$I = I_{1} + I_{2}$ $I=I_1+I_2$

$I = \frac{220}{R_{1}} + \frac{220}{R_{2}}$ $I=220/R_1+220/R_2$ [since $I = \frac{V}{R}$ $I=V/R$ ]

$I = (220) \times \frac{P_{1}}{V_{1}^{2}} + 220 \times \frac{P_{2}}{V_{2}^{2}}$ $I=(220)xxP_1/(V_1^2) +220xxP_2/V_2^2$ [since $R = \frac{V^{2}}{P}$ $R=V^2/P$ ]

where $P_{1}, V_{1} & P_{2}, V_{2}$ $P_1,V_1&P_2,V_2$ are ratings of bulbs respectively.

$I = (220) \times \frac{100}{220^{2}} + (220) \times \frac{25}{220^{2}}$ $I=(220)xx100/220^2+(220)xx25/220^2$

$I = 0.4545 + 0.1136$ $I=0.4545+0.1136$

$I = 0.5681 a m p s$ $I=0.5681 amps$

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Question #0ad4c

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