A 2-ohm and a 6ohm resistor are connected in series to a battery with V = 12 V. Calculate the power dissipated?

Question

A 2-ohm and a 6-ohm resistor are connected in series to a battery with V = 12 V. Calculate the power dissipated in the 6-ohm resistor.

A) 9 W
B) 12 W
C) 13.5 W
D) 2.67 W
E) Noneofthese.

SOMEONE PLEASE EXPLAIN!!!

Answer 1

Total resistance in series is R1+R2+R3 etc. so your total resistance is 2+6 = 8 ohms.

Power ( in watts, P) =Current (in amps, I) x Electromotive Force (in volts, EMF).

You have your EMF of 12 volts, but you need to find the current.

EMF = Current x Resistance (E= I x R, Ohm's law). So 12 = current x 8. 12/8 = 1.5, so your current is 1.5 amps.

Plug this in to figure out the total power: Power = 1.5 x 12. Total Power dissipated is 18 watts. But you only want the power dissipated on the 6 ohm resistor, so...

In a series circuit, the current remains constant through the circuit. The voltage drops in a series circuit as it meets resistance. The voltage drop across the 2 ohm resistor is E=1.5x2 --> 3 volts. The voltage drop across the 6 ohm resistor is E=1.5x6 --> 9 volts. The total voltage lost across resistors in a circuit will always equal the source voltage.

The power dissipated across the 6 ohm resistor is P=1.5x9 --> 13.5 watts.



I hope this helps. Let me know if I can clarify something. I've got some good visualizations you can use to understand it better.

Answer 2

The battery is without difficulty connected to an 18 ohm resistor. V=IR, 36=I(18) So the present out of the battery is 2 amps. Now don't forget the 12 ohm resistor. P=I^2R = (2)^2*12=forty eight W. You might be proper if the drawback is asking in regards to the vigour dissipated by means of the 12 ohm resistor. The 6 ohm resistor would dissipate 24W. Books can also be incorrect, simply seem on the Bible : ) If both resistors were related to the battery in parallel, their amazing resistance would be four ohms. P=V^2/R = 36^2/6 = 216W. You're proper. P=I(I)R might be used for this situation too, but you must calculate I by way of the 6 ohm resistor. The current by means of each and every resistor is exceptional.) P=VV/R could be used for the first obstacle, but you have got to calculate V across just the 12 ohm resistor first.

Answer 3

i = V/ΣR = 12/8 = 1.5 amp



P6 = i²*R = 1.5²*6 = 13.5 W