Verified answer

a) 7+14+19= 40 The total resistance of resistors in series is equal to the sum of their individual resistances

b) V=IR so I=V/R I=6/40 current will be the same in a series circuit.

c) P=I^2Rso P= 0.15^2 x 19 = .427 ( unless i misunderstand i think you are asking for power)

Hi!

When resistors are in series we need to make the sum of them to have the equivalent resistor.

The current in each resistor is the same so you simply compute the current in the equivalent resistor and that gives you the current in each resistor.

The power dissipated in each resistor is the product of the voltage across is times the current that flow in it. BUT, since by the Ohm's lay you know that that "Voltage" is "Resistor times Current", the power can be compute by R * I * I (R and current squared).

So in your examples:

a) RTot=7+14+19=40 ohms

b) Current = V / R = 6 / 40 = 0.15A or 150mA

c) Power would be R*I*I so 19 * 0.15 * 0.15 = 0.4275 Watts, so 427.5 mW.

Hope this helped.

Dennis.

SS got A and B right, but for C you're only looking for the power delivered to the one resistor.

P=(I^2)R = (0.15)^2 * 19 = 0.4275W

Sequence resistors just accumulate (add collectively) on the grounds that the current has to battle its method through one after which a different. The present will be the identical via all resistors, considering there is just one direction. So the present, I = V/R, the place R is solely the sum.