An RL circuit with two inductors of self-inductance l1 = 4. 0 H connected in par

Question

An RL circuit with two inductors of self-inductance l1 = 4. 0 H connected in parallel to each other, and connected to a resistor of 6. 0 ohm and a battery of 3. 0 V (Fig. 31. 43) Assume that the induction have no mutual inductance and resistance. When the battery is suddenly connected, what is the initial of change of the current in each inductor? What is the final, steady current in the resistor? What arc ' ready currents in each inductor? (Hint: The current in the inductors are inversely proportional to their cs. Why?)

Explanation / Answer

(a)the two inductors are connected in parallel therefore weget (1/L) = (1/L1) + (1/L2) = (1/4.0) +(1/2.0) = (1 + 2/4.0) = (3/4) or L = (4/3) H the total inductance in the circuit is in parallel to theresistor in the circuit.Therefore,the current flowing through theresistor R and inductor L is I = (E/(R2 + (wL)2)1/2) E = 3.0 V,R = 6.0 and w = 2f where f = 0 or w = 2 * 0 = 0 or I = (E/(R2 + (0 * L)2)1/2)= (E/R) or I = (3.0/6.0) = 0.5 A as the source of emf is a battery the potential differenceacross the resistor and inductance is same. when the battery is suddenly connected,the initial rate ofchange of current in each inductor is E = L1 * (dI1/dt) or (dI1/dt) = (E/L1) E = 3.0 V and L1 = 4.0 H similarly,we have (dI2/dt) = (E/L2) L2 = 2.0 H (b)the final steady state current in the resistor and the twoinductors is same.This is because the resistor and the combinationof the two inductors (L) are connected in series.The final steadystate current is I = 0.5 A.

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