Two inductors of 8 H and 2 H connected in series opposing with mutual inductance of 1 H will have equivalent:
8 H
To find the equivalent inductance of two inductors connected in series opposing with mutual inductance, we use the following formula:
L_{\text{eq}} = L_1 + L_2 - 2M
where:
- L_1 is the inductance of the first inductor (8 H).
- L_2 is the inductance of the second inductor (2 H).
- M is the mutual inductance between them (1 H).
Substitute the values into the formula:
L_{\text{eq}} = 8 + 2 - 2 \times 1
Simplify the expression:
L_{\text{eq}} = 8 + 2 - 2 = 8 \, \text{H}
Therefore, the equivalent inductance is 8 H.
Let's verify the options:
- 8 H - This is the correct answer.
- 9 H - Incorrect, evaluated value is 8 H.
- 5 H - Incorrect, does not match the calculation.
- 11 H - Incorrect, does not match the calculation.
This confirms that the correct answer is 8 H.
Important Questions from Mutual Inductance
Two inductors whose self-inductance is 80 mH and 60 mH are connected in parallel aiding. The equivalent inductance of the combination is 48.75 mH. Calculate their mutual inductance.
Two inductors L1 = 20 mH and L2 = 40 mH are connected in series so that their equivalent inductance is 50 mH. The mutual inductance between the two coils is _______.
A transformer has 350 primary turns and 1050 secondary turns. The primary winding is connected across a 230 V, 50 Hz supply. The induced EMF in the secondary will be
Which of the following devices does not work on the principle of mutual induction?
Which statement is INCORRECT for mutual inductance?
