Three resistors are connected in series across 10V power supply as shown. The first resistor has a value of 1 Ohm, the second has a voltage drop of 1V, and the third has a power dissipation of 1W. Find the value of the circuit current. There are two possible answers. How do I figure this out?

 

Post your answers.

 

i get two answers. since it is a series circuit current is constant.

using R3 I=P/E 1W/10V=.1A
using R2 10V/.1A= 10 Ohm
VR2 is 1V/10 Ohm = .1A
using R1 10V/1 Ohm is 10A

 

using R3 I=P/E 1W/10V=.1A
using R2 10V/.1A= 10 Ohm
VR2 is 1V/10 Ohm = .1A
using R1 10V/1 Ohm is 10A

Why must you have two answers? Is that what the instructor stated?

Can you tell me about Kirchoff's Voltage Law as it applies to your circuit?

 

cgeorge5150,

using R3 I=P/E 1W/10V=.1A

But 10 volts is not dropped only across R3, so that is wrong.

using R2 10V/.1A= 10 Ohm

Wrong amps from first calculation, so wrong for second calculation.

VR2 is 1V/10 Ohm = .1A

Wrong on second calculation, so wrong on third calculation.

using R1 10V/1 Ohm is 10A

Wrong because 10 volts is spread across all three resistors

OK, lets get down to what works. Who cares what the two unknown resistors are? We are trying to find current. Looks to me like the best way to do that is to equate the power from the 10 volt source to the power dissipated by the resistors. So let's do that.

T = total resistance of all three resistors, then I = 10/T

The general equation is E^2/R = R*I^2 + I*V + 1

The particular equation is 10^2/T = 1*(10/T)^2 + 1*10/T + 1

Solving the quadratic gives T = 1.125178063, and T = 88.87482194

So I = 10/1.125178063 = 8.887482194 amps
and I = 10/88.87482194 = 0.1125178063 amps

Ratch