find the current in 10 ohm resistance using superposition theorem
Answers
Explanation:
1 Answer +1 vote answered Sep 21, 2019 by Abhinav03 (64.6k points) selected Sep 21, 2019 by Vikash Kumar 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.