Test: Combination of Resistors - NEET MCQ

# Test: Combination of Resistors - NEET MCQ

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## 5 Questions MCQ Test Physics Class 12 - Test: Combination of Resistors

Test: Combination of Resistors for NEET 2024 is part of Physics Class 12 preparation. The Test: Combination of Resistors questions and answers have been prepared according to the NEET exam syllabus.The Test: Combination of Resistors MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Combination of Resistors below.
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Test: Combination of Resistors - Question 1

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

Detailed Solution for Test: Combination of Resistors - Question 1

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

Test: Combination of Resistors - Question 2

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

Detailed Solution for Test: Combination of Resistors - Question 2

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/Req​​=(1/2+3)​+(1/4+6)​
1/Req​​=(1/5)+(1/10)
∴Req​=(5×10/15)​
∴Req​=(10​/3) Ω

Test: Combination of Resistors - 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

Detailed Solution for Test: Combination of Resistors - Question 3

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.

Test: Combination of Resistors - 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?

Detailed Solution for Test: Combination of Resistors - Question 4

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​=R2​R1
​​⇒1/9​= R2​/90
​⇒R2​=90/9​=10
∴R2​=shunt=10Ω

Test: Combination of Resistors - Question 5

Calculate equivalent resistance when three resistances of 2 ohms, 3 ohms and 6 ohms are connected in parallel combination.

Detailed Solution for Test: Combination of Resistors - Question 5

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