Test: Internal Resistance & Cells in Series and in Parallel (December 5) - NEET MCQ

# Test: Internal Resistance & Cells in Series and in Parallel (December 5) - NEET MCQ

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## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Internal Resistance & Cells in Series and in Parallel (December 5)

Test: Internal Resistance & Cells in Series and in Parallel (December 5) for NEET 2024 is part of Daily Test for NEET Preparation preparation. The Test: Internal Resistance & Cells in Series and in Parallel (December 5) questions and answers have been prepared according to the NEET exam syllabus.The Test: Internal Resistance & Cells in Series and in Parallel (December 5) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Internal Resistance & Cells in Series and in Parallel (December 5) below.
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Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 1

### A battery of emf 15V and internal resistance of 4Ω is connected to a resistor. If the current in the circuit is 2A and the circuit is closed. Resistance of the resistor and terminal voltage of the battery will be

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 1

Given, ε = 15V,
r = 4Ω,
I = 2A
Now, for resistance of the resistors
ε − Ir = V = IR;
15 − 2 × 4=2 × R;
15 − 8  = 2R
R = 7/2 = 3.5Ω.
Terminal voltage of battery,
V = IR = 2 × 3.5 = 7V

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 2

### The total resistance in the parallel combination of three resistances 9Ω, 7Ω and 5Ω

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 2

In the parallel combination of three resistance's resistance is

or

= 2.02Ω

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Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 3

### Equivalent resistance (in ohm) of the given network is

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 3

Between points A and B all resistances are combined in series
∴ Req = 3Ω + 4Ω + 5Ω + 6Ω
= 18Ω
Between points A and B all resistances are combined in series
∴ Req = 3Ω + 4Ω + 5Ω + 6Ω
= 18Ω

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 4

Equivalent resistance of the given network between points A and B is

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 4

In each segment of the combination 3Ω and 2Ω resistance are connected in series separately.
∴ R' = +3 = 6Ω and R'' = 2 + 2 = 4Ω
R' and R'' are connected in parallel
∴ For first segment

Similarly for second and third segment

Now segment is connected in series then the total resistance of combination is
Req = Req1 + Req2 + Req3

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 5

Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. This combination is connected to a battery of emf 20 V and negligible internal resistance, the total current drawn from the battery is

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 5

Potential of 20V20V will be same across each resistance current

∴ Total current drawn from circuit

I = I1 + I2 + I3

= 10 + 5 + 4 = 19A

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 6

n resistors each of resistance R first combine to give maximum effective resistgance and then combine to give minimum. The ratio of the maximum resistance is

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 6

To get maximum equivalent resistance all resistances must be connected in series
∴ (Req)max = R + R + R...ntimes = nR
To get minimum equivalent resistance all resistances myst be connected in parallel.
∴

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 7

A battery having 12V emf and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 1A, then the resistance of resistor and lost voltage of the battery when circuit is closed will be

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 7

Here,
ε = 12V,  r = 3Ω, I = 1A, V = IR = ε − Ir
∴
= 12 − 3 = 9Ω
and V = IR = 1 × 9 = 9V

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 8

A copper cylindrical tube has inner radius a and outer radius b. The resistivity is ρ. The resistance of the cylinder between the two ends is

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 8

If one had considered a solid cylinder of radius b, one can suppose that it is made of two concentric cylinders of radius a and the outer part, joined along the length concentrically one inside the other.
If Ia and Ix are the currents flowing through the inner and outer cylinders
∵ Itotal = Ib = Ia + Ix
⇒ VRb = V/Ra + V/Rx
where Rb is the total resistance and Rx is the resistance of the tubular part.
∴
But
∴
∴

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 9

A wire of resistance 12 ohms per meter is bent to form a complete circle of radius 10cm. The resistance between its two diametrically opposite points, A and B as shown in the figure is

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 9

Wire of length 2π x 0.1 m of 12Ωm-1 is bent to a circle.

Resistance of each part = 12 x π x 0.1
= 1.2π Ω
Total resistance = 0.6π Ω

Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 10

Combine three resistors 5Ω, 4.5Ω and 3Ω in such a way that the total resistance of this combination is maximum

Detailed Solution for Test: Internal Resistance & Cells in Series and in Parallel (December 5) - Question 10

For maximum equivalent or total resistance of the resistors must be combined in series.
Req = R+ R2 = R3 = 5 + 4.5 + 3
= 12.5Ω

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