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Two resistors, one with 7.00 Ω of resistance and the other with 11.00 Ω of resistance, are connected in series to a 9.00 V battery. How much power is dissipated by these two resistors?

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asuwafohjames

Answer:

4.5 W

Explanation:

Applying,

P = V²/(R₁+R₂).................. Equation 1

Where P = Power, V = Voltage, R₁ and R₂ = values of the two resistor.

From the question,

Given: V = 9.00 V, R₁ = 7.00 Ω, R₂ = 11.00  Ω

Substitute these values into equation 1

P = 9²/(7+11)

P = 81/(18)

P = 4.5 Watt.

Hence the power dessipated by the two resistors is 4.5 watt

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