What is the total power dissipated by all four 100 Ω resistors in series across a 40 V supply?

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Study for the BCTC Industrial Maintenance Technology AMTEC – NOCTI Mechatronic Assessment. Prepare with comprehensive question sets and detailed explanations. Ensure success with our targeted practice!

Multiple Choice

What is the total power dissipated by all four 100 Ω resistors in series across a 40 V supply?

Explanation:
In a series circuit, the same current flows through every resistor and the resistors’ resistances add up. Four 100 Ω resistors in series give a total of 400 Ω. The current from a 40 V supply is I = V / R_total = 40 / 400 = 0.1 A. The power dissipated by a resistor is P = I^2 R, so each resistor uses P = (0.1)^2 × 100 = 1 W. With four resistors, the total power is 4 W. You can also get the same result with P_total = V^2 / R_total = 40^2 / 400 = 4 W. Therefore, the total power dissipated is 4 W.

In a series circuit, the same current flows through every resistor and the resistors’ resistances add up. Four 100 Ω resistors in series give a total of 400 Ω. The current from a 40 V supply is I = V / R_total = 40 / 400 = 0.1 A. The power dissipated by a resistor is P = I^2 R, so each resistor uses P = (0.1)^2 × 100 = 1 W. With four resistors, the total power is 4 W. You can also get the same result with P_total = V^2 / R_total = 40^2 / 400 = 4 W. Therefore, the total power dissipated is 4 W.

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