Combination of Inductors
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Q. Pure inductance of 3.0 H is connected as shown below. The equivalent inductance of the circuit is
- 2 H
- 3 H
- 9 H
- 1 H
Q. The equivalent inductance of two dissimilar inductors is 2.4 H, when connected in parallel and 10 H, when connected in series. The inductance of one of the inductors is -
- 2 H
- 6 H
- 5 H
- 7 H
Q. Two inductances connected in parallel are equivalent to a single inductance of 1.5 H and when connected in series are equivalent to a single inductance of 8 H. The difference in their inductance is-
- 3 H
- 7.5 H
- 2 H
- 4 H
Q. Three pure inductances each of 3 H are connected as shown in fig. The equivalent inductance between points A and B is
- 1 H
- 2 H
- 3 H
- 9 H
Q. Two inductors 0.4 H and 20.6 H are connected in parallel. If this combination is connected in series with an inductor of inductance 0.76 H. The equivalent inductance of the circuit will be (nearly)
- 2 H
- 0.1 H
- 0.2 H
- 1 H
Q. A circuit contains two inductors of self-inductance L1 and L2 in series (see figure). If M is the mutual inductance, then the effective inductance of the circuit shown will be
- L1+L2−2M
- L1+L2+M
- L1+L2+2M
- L1+L2
Q. In the following digital circuit, what will be the output at ′Z′, when the input (A, B) are (1, 0), are (1, 0), (0, 0), (1, 1), (0, 1)
- 0, 0, 1, 0
- 1, 0, 1, 1
- 1, 1, 0, 1
- 0, 1, 0, 0
Q. Find the equivalent inductance in the combination given below between points A and B.
- 3.8 mH
- 3.1 mH
- 2.4 mH
- 5.6 mH
Q. The equivalent inductance of the network shown in the figure, between the points a and b is -
- 4H3
- 2H3
- H3
- H
Q. Pure inductors each of inductance 3 H are connected as shown. The equivalent inductance of the circuit is
- 1 H
- 2 H
- 3 H
- 9 H
Q. Find the equivalent resistance between A and E (Resistance of each resistor is R).
- 712R
- 713R
- 715R
- 813R
Q. A circuit contains two inductors of self-inductance L1 and L2 in series as shown in the figure. If M is the mutual inductance, then the effective inductance of the circuit shown will be
- L1+L2−M
- L1+L2+3M
- L1+L2+2M
- L1+L2
Q. Two inductors 0.4 H and 20.6 H are connected in parallel. If this combination is connected in series with an inductor of inductance 0.76 H. The equivalent inductance of the circuit will be (nearly)
- 2 H
- 1 H
- 0.2 H
- 0.1 H
Q. In which of the following circuits, is the current maximum, just after the switch (S) is closed?
- (i)
- (ii)
- (iii)
- Data insufficient
Q. Two capacitors C1 and C2 are connected in series, assume that C1<C2. The equivalent capacitance of this arrangement is C, where
- C1<C<C2
- C<C1/2
- C1/2<C<C1
- C2<C<2C2
Q. Six equal capacitors each of capacitance C are connected as shown in the figure. The equivalent capacitance between points A and B is :
- 1.5 C
- C
- 0.5C
- 2C
Q. Pure inductance of 3.0 H is connected as shown below. The equivalent inductance of the circuit is
- 1 H
- 2 H
- 3 H
- 9 H
Q. Three pure inductances each of 3 H are connected as shown in fig. The equivalent inductance between points A and B is
- 1 H
- 2 H
- 3 H
- 9 H
Q. Find the equivalent capacitance across AB (all capacitances in μ F) :
- 203μF
- 9μF
- 48μF
- None of these
Q. Pure inductance of 3.0 H is connected as shown below. The equivalent inductance of the circuit is
- 1 H
- 2 H
- 3 H
- 9 H
Q. The inductor arrangement shown in the figure, with L1=50mH, L2=80mH, L3=20mH and L4=15mH, is to be connected to varying current source. What is the equivalent inductance of the arrangement?
- 18mH
- 71mH
- 55mH
- 81mH
Q.
The equivalent capacity between the points ‘A’ and ‘B’ in the following figure will be:
- 9μF
- 1μF
- 4.5μF
- 6μF
Q. Seven capacitors, each of capacitance 2μF are to be connected to obtain a capacitance of 10/11μF. Which of the following combinations is possible?
- 3 in parallel 4 in series
- 2 in parallel 5 in series
- 5 in parallel 2 in series
- 4 in parallel 3 in series
Q. The magnetic potential energy stored in a certain inductor is 25 mJ, When the current in the inductor is 60 mA This inductor is of inductance-
- 0.138 H
- 138.88 H
- 1.389 H
- 13.89 H
Q. An enquiring physics student connects a cell to a circuit and measures the current drawn from the cell to I1. When he joins a second identical cell in series with the first, the current becomes I2. When the cells are connected in parallel, the current through the circuit is I3. Show that relation between the current is 3I3I2=2I1(I2+I3).
Q. The equivalent inductance of two inductors is 2.4H when connected in parallel and 10 H when connected in series. What is the value of inductances of the individual inductors?
- 7H, 3H
- 6H, 4H
- 5H, 5H
- 8H, 2H
Q. Three pure inductance are connected as shown in the Fig.26.70. The equivalent inductance of the circuit is
- 2.25 H
- 2.75 H
- 2.00 H
- 1.75 H
Q.
Two inductors each equal to L are joined in parallel. The equivalent inductance is
- L
- 2L
- zero
- L2
Q. To get an output Y=1 from the circuit shown, the input A, B and C must be respectively.
- 0, 1, 0
- 1, 1, 0
- 1, 0, 0
- 1, 0, 1
Q. This circuit is equivalent to a single inductor having inductance Leq =18H . Determine the value of the inductance L.
- 20H
- 40H
- 60H
- 80H