B.Tech EEE — free MCQ practice

1. Series resistances add, so the three values go together directly. 2. Rs = 2.0 ohm + 4.0 ohm + 6.0 ohm = 12.0 ohm. 3. The value 1.1 ohm is what the same three resistors give in parallel, which is always below the smallest of them. 4. Keeping only 6.0 ohm or averaging to 4.0 ohm ignores that every resistor in the string is met by the same current. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.1 Resistors in Series and Parallel_

1. In series the resistances add: Rs = 1.00 ohm + 6.00 ohm + 13.0 ohm = 20.0 ohm. 2. Ohm's law for the whole loop gives I = V / Rs. 3. I = 12.0 V / 20.0 ohm = 0.600 A. 4. That one value of 0.600 A flows through all three resistors, since a series string offers a single path. 5. The figure 14.9 A is what the same battery would deliver if the three resistors were in parallel, so it is the trap for misreading the connection. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.1 Resistors in Series and Parallel_

1. Power in a resistor is P = IV, and substituting V = IR turns this into P = I squared times R. 2. Every element of a series string carries the same current, here 0.600 A. 3. I squared is (0.600 A)(0.600 A) = 0.360 A squared. 4. For the 6.00 ohm resistor, P = (0.360 A squared)(6.00 ohm) = 2.16 W. 5. As a check, the three resistors dissipate 0.360 W, 2.16 W and 4.68 W, adding to 7.20 W, which equals the battery output of (0.600 A)(12.0 V). 6. 0.360 W is the power in the 1.00 ohm resistor, picked up by using the wrong resistance in the same formula. 7. 4.68 W is the power in the 13.0 ohm resistor, the largest share because the current is common and the resistance is largest. 8. 24.0 W is what this resistor would dissipate with the full 12.0 V across it, which is the parallel case, not the series one. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.1 Resistors in Series and Parallel_