What is current I in the circuit as shown in figure?
Three 2Ω resistors are in series. Their total resistance =6Ω. Now it is in parallel with 2Ω resistor, so total resistance,
Five resistances are connected as shown in fig. The effective resistance between points A and B is
In the given circuit, the arrangement of resistances is a form of Wheatstone's bridge. Hence no current will flow through the 7Ω resistor. The resistances between A and B will be a parallel combination of each other.
Three resistors of 4Ω, 12Ω , and 6Ω are connected in parallel. No. of 12Ω resistors required to be connected in parallel to reduce the total resistance to half of its original is
Here 4 Ω, 12 Ω, 6 Ω when connected in parallel results in 2Ω. to reduce it to half we have to join 1\R original = 6\12 for reducing it to half we have to join 6 , 12 Ω resistors in parallel (6\12) + (1\ 12 × 6) = 12\12 = 1 ohm . Half of its original value therefore option a is correct.
The current in a coil of resistance 90 ohms is to be reduced by 90 percent. What value of resistance should be connected in parallel with it?
We know that,
∴I1/I2 = R2/ R1
Current flowing through 90Ω resistance is reduced by 90%
∴ Current ratio =10%:90%
The equivalent resistance in a parallel resistance circuit when resistances are 2 Ω, 3 Ω and 6 Ω, and V = 6 V is
If total resistance of wire = 6 ohm.
Then each side of an equilateral triangle has 6/3 = 2 ohm.
If we choose BC.
Then AB and AC are in series with BC..
Hence 1/R = 1/2 + 1/4
=) 1/R = 3/4 ohm
=) R = 4/3 ohm.