Alternating Current Test

Result

Alternating Current Test - 3

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Question 1
1 / -0

If two inductances are joined in parallel, the equivalent inductance is 2.4 H, and it becomes 10 H, if they are joined in series. Then, the difference between both of them will be

A
2 H

B
10 H

C
5 H

D
4 H

Solution

In series, L₁+ L₂ = 10
In parallel, L₁L₂/L₁+L₂ = 2.4
So, L₁L₂ = 24
(L₁- L₂)² = (L₁+ L₂)² - 4L₁L₂= 100 - 96 = 4
L₁- L₂= 2 H
Question 2
1 / -0

The magnitude of self induced emf in a coil of 0.4 Henry self inductance, when current in it is changing at the rate of 500 A/s, is

A
8 x 10^-4V

B
8 x 10^-3V

C
200 V

D
500 V

Solution

Magnitude of the emf induced is
e = L × dl/dt
= 0.4 × 500 = 200 V
Question 3
1 / -0

Directions For Questions

In the following question, a statement of Assertion is given, followed by a corresponding statement of Reason. Mark the correct answer as

(a) if both Assertion and Reason are true and the Reason is the correct explanation of the Assertion
(b) if both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion
(c) if Assertion is true, but Reason is false
(d) if both Assertion and Reason are false

...view full instructions

Assertion: In the phenomenon of mutual induction, self-induction of each of the coils persists.
Reason: Self-induction arises when strength of current in one coil changes. In mutual induction, current changes in individual coils.

A
(a)

B
(b)

C
(c)

D
(d)

Solution

Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. In the case of self-inductance, the magnetic field created by a changing current in the circuit itself induces a voltage in the same circuit. Therefore, the voltage is self-induced.

When the ﬂux from one coil cuts another adjacent (or magnetically coupled) coil, an emf is induced in the second coil. In accordance with the Lenz's law, the emf induced in the second coil sets up a ﬂux that opposes the original ﬂux from the ﬁrst coil. Thus, the induced emf is again a counter-emf.
Question 4
1 / -0

If 8 Ω resistance and 6 Ω reactance are present in an AC series circuit, what will be the impedance of the circuit?

A
20 ohm

B
5 ohm

C
10 ohm

D
14√2 ohm

Solution

Z = (R² + X_c²)^1/2
Z = (64 + 36)^1/2 = 10 ohm