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 _______.
Which statement is INCORRECT for mutual inductance?
Calculate the total inductance (in H) of the circuit shown below:
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
