In the circuit shown in figure, key K is closed at t=0, the current through the key at the instant t=10-3 ln 2 s is

Question

In the circuit shown in figure, key K is closed at t=0, the current through the key at the instant t=10-3 ln 2 s is (A) 2 A (B) 3.5 A (C) 2.5 A (D) 0

Tardigrade Admin · Accepted Answer

Current in branches containing L and R will flow independently I1=(20/10)(1-e-(t/5 × 10-4)) =(3/2)=1.5 A <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/e9e5e1354a4a4bbcdd4aa4269e7831e5-.png" /> I2=(20/10) e-(t/10-3)=1.0 A I=I1+I2=2.5 A

Q. In the circuit shown in figure, key $K$ is closed at $t = 0$ , the current through the key at the instant $t = 1 0^{- 3} ln 2 s$ is

2 A

3.5 A

2.5 A

0

Current in branches containing $L$ and $R$ will flow independently
$I_{1} = \frac{20}{10} (1 - e^{- \frac{t}{5 \times 1 0 ^{- 4}}})$
$= \frac{3}{2} = 1.5 A$

$I_{2} = \frac{20}{10} e^{- \frac{t}{1 0 ^{- 3}}} = 1.0 A$
$I = I_{1} + I_{2} = 2.5 A$

Q. In the circuit shown in figure, key K is closed at t=0, the current through the key at the instant t=10−3ln2s is

2 A

3.5 A

2.5 A

0

Current in branches containing L and R will flow independently I1​=1020​(1−e−5×10−4t​) =23​=1.5A I2​=1020​e−10−3t​=1.0A I=I1​+I2​=2.5A

Q. In the circuit shown in figure, key $K$ is closed at $t = 0$ , the current through the key at the instant $t = 1 0^{- 3} ln 2 s$ is

Current in branches containing $L$ and $R$ will flow independently
$I_{1} = \frac{20}{10} (1 - e^{- \frac{t}{5 \times 1 0 ^{- 4}}})$
$= \frac{3}{2} = 1.5 A$

$I_{2} = \frac{20}{10} e^{- \frac{t}{1 0 ^{- 3}}} = 1.0 A$
$I = I_{1} + I_{2} = 2.5 A$