![]() ![]() Superposition principle is applied in the circuit of figure 12 taking on source at a time as shown in figure 13. Obtain the steady state current through the 10V battery in time domain in the circuit of figure 12 using superposition theorem. Thus the ratio of voltages is 0.89 ∠-26.57 o only. In figure 11(b), Z’’ (impedance across V 2) Where, Z’ is the impedance of the network across V 1 Principle of superposition is applied in the network taking each voltage source at a time (figure 11(a) and 11(b)). What is the value of the ratio of the two sources? In the network of figure 10 the current flowing through the 5Ω resistor is equal when each of the voltages sources acts separately on the circuit. ⸫ I 1 the total current through R L is I = I 1’ + I 1’’ (since both are directed in the same direction) ![]() Principle of superposition is applied in the given circuit taking each source at a time (figure 8 and 9). Thus, the current through 5Ω resistor, using the principle of superposition isįind the current in the resistor (R L) using the principle of superposition in figure 7. Superposition theorem is applied to the given network as shown in figure 5 and 6 deactivating one source at a time. ⸫ I 2 (following the principle of superposition),įind, by the principle of superposition, the current through 5Ω resistor (figure 4). Using the principle of superposition, the given circuit is fed by each of the two sources at a time (figure 2 & 3). ![]() Find the current through j3Ω inductive reactance using the principle of superposition. ![]()
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