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QUESTION: 1

What is current I in the circuit as shown in figure?

Solution:

Three 2Ω resistors are in series. Their total resistance =6Ω. Now it is in parallel with 2Ω resistor, so total resistance,

1/R=1/2+1/6=3+1/6=4/6=2/3

R=3/2

∴I=RV=3/(3/2)=3×2/3=2A

QUESTION: 2

Five resistances are connected as shown in fig. The effective resistance between points A and B is

Solution:

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.

Hence,

1/R_{eq}=(1/2+3)+(1/4+6)

1/R_{eq}=(1/5)+(1/10)

∴R_{eq}=(5×10/15)

∴R_{eq}=(10/3) Ω

QUESTION: 3

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

Solution:

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.

QUESTION: 4

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?

Solution:

We know that,

I=V/R

∴I_{1}/I_{2} = R_{2}/ R_{1}

R_{1}=90Ω

Current flowing through 90Ω resistance is reduced by 90%

∴ Current ratio =10%:90%

=1:9

∴1/9=R_{2}R_{1}

⇒1/9= R_{2}/90

⇒R_{2}=90/9=10

∴R_{2}=shunt=10Ω

QUESTION: 5

The equivalent resistance in a parallel resistance circuit when resistances are 2 Ω, 3 Ω and 6 Ω, and V = 6 V is

Solution:

If total resistance of wire = 6 ohm.

Then each side of an equilateral triangle has 6/3 = 2 ohm.

Diagram :

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.

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