Electric Current Test

Result

Electric Current Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

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

A
1 A

B
2.0 A

C
1.2 A

D
0.5 A

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
1 / -0

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

A
20/30Ω

B
15Ω

C
10/3Ω

D
6Ω

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
1 / -0

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

A
6

B
3

C
12

D
2

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
1 / -0

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?

A
100Ω

B
10Ω

C
9Ω

D
90Ω

Solution

We know that,
I=V/R
∴I₁/I₂ = R₂/ R₁
R₁=90Ω
Current flowing through 90Ω resistance is reduced by 90%
∴ Current ratio =10%:90%
=1:9
∴1/9=R₂R₁
⇒1/9= R₂/90
⇒R₂=90/9=10
∴R₂=shunt=10Ω
Question 5
1 / -0

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

A
2 Ω

B
3 Ω

C
1 Ω

D
4 Ω

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.