In the circuit shown below, the key K is closed at t = 0. The current through the battery is:
VR1R2√R21+R22 at t=0 and VR2 at t=∞
VR2 at t=0 and V(R1+R2)R1R2 at t=∞
VR2 at t=0 and VR1R2√R21+R22 at t=∞
V(R1+R2)R1R2 at t=0 and VR2 at t=∞
A
VR1R2√R21+R22 at t=0 and VR2 at t=∞
B
VR2 at t=0 and VR1R2√R21+R22 at t=∞
C
V(R1+R2)R1R2 at t=0 and VR2 at t=∞
D
VR2 at t=0 and V(R1+R2)R1R2 at t=∞
Open in App
Solution
Verified by Toppr
At t=0, inductor behaves like an infinite resistance
Was this answer helpful?
0
Similar Questions
Q1
In the circuit shown below, the key K is closed at t=0. The current through the battery is
View Solution
Q2
For a circuit shown in Figure switch S1, is closed at t=0, then at t=(2R2+R1)C,S1 is opened and S2 is closed. a. Find the charge on capacitor at t=(2R2+2R1)C. b. Find current through R2 (adjacent to the battery) at t=(3R1+2R2)C.
View Solution
Q3
What is the value of V2 (potential difference across R2) at t=0 and t=∞?
View Solution
Q4
In the circuit shown, key (K) is closed at t=0. Current through key at t=10−3ln(2) is
View Solution
Q5
A capacitor of capacity 6μF and initial charge 160μC is connected with a key s and resistance as shown in figure. Point M is earthed. If key is closed at t=0; then the current through resistances R1 and R2 at t=16μs are