In the circuit of figure 23, find the current through the inductor as well as through the capacitor using the principle of superposition. In figure 22, I 1” (current through the capacitor due to 100 ∠ 0 o V) = 0, since the terminals are seen to be shorted. Superposition principle is applied in the circuit as shown in figure 21 and 22 deactivating one source at a time. In figure 18 the impedance Z’ across 100 ∠0 o V isįind the current through the capacitor by superposition theorem (figure 20). Superposition is applied in the given circuit as shown in figure 18 and 19 deactivating each source at time. Thus, the current through the capacitor is 6.41 ∠119.117 o A.įind the current through the capacitor of (-j5)Ω reactance in figure 17 using Superposition Theorem. Let us apply superposition theorem (figure 15 and 16) deactivating one source at a time. Thus, the battery current, in time domain, isįind the current in the (-j6Ω) capacitive reactance using superposition theorem in the circuit of figure 14. ⸫ Net current through the resistor, using the principle of superposition, is In time domain, the instantaneous value is Where Z’’ is the impedance of the circuit shown in figure 13(b) across the a.c. Since j2Ω inductive reactance would act as short circuit while -j4Ω capacitive reactance would act as open circuit while encountering 10V d.c source, the currents I 1’ is equal to I’ and I 2’ = 0. 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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