what is the power loss in the 10ohm resistor. use Thevenins theorem
Answers
When the 10 Ω resistance is removed the circuit becomes as shown in Fig..Now we will find the open-circuit voltage VAB = Vth. For this purpose we will go from point B to point A and find the algebraic sum of the voltages met on the way. It should be noted that with terminals A and B open there is no voltage drop on the 8 Ω resistance. However the two resistances of 5 Ω and 2 Ω are connected in series across the 20-V battery. As per voltage-divider rule drop on 2 Ω resistance = 20 × 2/2 + 5 = 5.71 V with the polarity as shown in figure. As per the sign convention of Art. VAB = Vth = + 5.71 − 12 = − 6.29 V The negative sign shows that point A is negative with respect to point B or which is the same thing point B is positive with respect to point A. For finding RAB = Rth we replace the batteries by short-circuits as shown in Fig. . ∴ RAB = Rth = 8 + 2 || 5 = 9.43 Ω Hence the equivalent Thevenin’s source with respect to terminals A and B is as shown in Fig. When 10 Ω resistance is reconnected across A and B current through it is I = 6.24/9.43 + 10 = 0.32 A