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3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

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JIET Group of Institutions Colleges in Jodhpur

Contributor-Level 10

3.23 Internal resistance of the cell = r

Balance point of the cell in open circuit, $l_{1}$ = 76.3 cm

An external resistance R is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, $l_{2}$ = 64.8 cm

Current flowing through circuit = I

The relation connecting resistance and emf is,

r = ( $\frac{l_{1} - l_{2}}{l_{2}}$ )R = $\frac{76.3 - 64.8}{64.8}$ $\times 9.5 =$ 1.69 Ω

Therefore, the internal resistance is 1.69 Ω.

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11 months ago

Does JIET Group of Institutions provide 100% placements?

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Contributor-Level 10

The placement rate for JIET Group of Institutionsin 2024 was 75%. However, JIET Group of Institutions provides good placement to its students. The data tabulated below provides JIET Group of Institutions placement highlights for 2024:

Particulars	Placement Statistics (2024)
No. of Students Placed	213
No. of Companies Visited	200
Placement Rate	75%
the highest Package	INR 11 LPA

The table below provides the department wise average package recorded at JIET Group of Institutions:

Department	Average Package (2024)
AI ML	INR 5.35 LPA
ECE	INR 4.15 LPA
EE	INR 4.18 LPA
ME	INR 4.6 LPA
CE	INR 4.6 LPA
JIHM	INR 3.2 LPA
MBA	INR 5.32 LPA
DS	INR 5.21 LPA
CSE	INR 5.66 LPA

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11 months ago

3.22 Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf $?$ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is

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(a) What is the value e ?

(b) What purpose does the high resistance of 600 kΩ have?

(c) Is the balance point affected by this high resistance?

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

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JIET Group of Institutions JIET Group of Institutions Placement

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3.22 Constant emf of the standard cell, $E_{1}$ = 1.02 V

Balance point on the wire, $l_{1}$ = 67.3 cm

A cell of unknown emf, $?$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is, $\frac{E_{1}}{l_{1}}$ = $\frac{?}{l}$

Hence, $?$ = $\frac{l}{l_{1}} \times E_{1}$ = $\frac{82.3}{67.3} \times 1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

...more

3.22 Constant emf of the standard cell, $E_{1}$ = 1.02 V

Balance point on the wire, $l_{1}$ = 67.3 cm

A cell of unknown emf, $?$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is, $\frac{E_{1}}{l_{1}}$ = $\frac{?}{l}$

Hence, $?$ = $\frac{l}{l_{1}} \times E_{1}$ = $\frac{82.3}{67.3} \times 1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

The method would not work if the emf of the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

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11 months ago

How to crack JIET Group of Institutions placement interview?

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Contributor-Level 10

To crack JIET Group of Institutions placement interview students can follow the steps given below:

· Review your resume

· Practice common interview questions

· Showcase your strengths

· Be confident and articulate

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11 months ago

3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

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JIET Group of Institutions Colleges in Rajasthan

Contributor-Level 10

3.21 Let the equivalent resistance of the given circuit be R’. The equivalent resistance of an infinite network is given by

R’ = 2 + $\frac{R'}{(R^{'} + 1)}$ or R’ = $\frac{2 R^{'} + 2 + R'}{(R^{'} + 1)}$ or ${R'}^{2}$ + R’ = 2R’ + 2 + R’

${R'}^{2} - 2 R^{'} - 2 = 0$

R’ = $\frac{2 \pm \sqrt{4 + 8}}{2}$ = 1 $\pm \sqrt 3$

Since R’ cannot be negative, hence R’ = 1+ $\sqrt 3$ = 2.73 Ω

Internal resistance, r = 0.5Ω

Total resistance = 2.73 + 0.5 = 3.23 Ω

Current drawn from the source = $\frac{12}{3.23}$ A= 3.72 A

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11 months ago

How are the MBA placements at JIET Group of Institutions?

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Contributor-Level 10

Course wise placements for JIET Group of Institutions has not been released yet. However, the average package for MBA stood at INR 5.32 LPA. Students can refer to the key highlights of the JIET Group of Institutions placement statistics given in the table below:

Particulars	Placement Statistics (2024)
No. of Students Placed	213
No. of Companies Visited	200
Placement Rate	75%
the highest Package	INR 11 LPA

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11 months ago

3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

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JIET Group of Institutions Colleges in Jodhpur

Contributor-Level 10

3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{e f f}$ is maximum. $R_{e f f}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{e f f}$ is minimum. $R_{e f f}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{n R}{\frac{R}{n}}$ = $n^{2}$

Let us assume, $R_{1}$ = 1 Ω, $R_{2}$ = 2 Ω, $R_{3}$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$ Ω

From the circuit the equivalent resistance is given by

R = $\frac{2 \times 1}{2 + 1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

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3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{e f f}$ is maximum. $R_{e f f}$ = nR

Let us assume, $R_{1}$ = 1 Ω, $R_{2}$ = 2 Ω, $R_{3}$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$ Ω

From the circuit the equivalent resistance is given by

R = $\frac{2 \times 1}{2 + 1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{2 \times 3}{2 + 3}$ + 1 = $\frac{6}{5}$ + 1 = $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

Required equivalent resistance, R = $\frac{6}{11}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{1 \times 2 \times 3}{1 \times 2 + 2 \times 3 + 3 \times 1}$ = $\frac{6}{11}$ Ω

(a)

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2 $Ω$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R = $\frac{2 \times 4}{2 + 4}$ = $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is $\frac{4}{3}$ Ω $\times 4 = \frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5 $\times R = 5 R$

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11 months ago

How are the average packages at JIET Group of Institutions?

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University College London Imperial College

Contributor-Level 10

The department wise average packages recorded during JIET Group of Institutions placements 2024 are presented in the table below:

Department	Average Package (2024)
AI ML	INR 5.35 LPA
ECE	INR 4.15 LPA
EE	INR 4.18 LPA
ME	INR 4.6 LPA
CE	INR 4.6 LPA
JIHM	INR 3.2 LPA
MBA	INR 5.32 LPA
DS	INR 5.21 LPA
CSE	INR 5.66 LPA

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11 months ago

Which has higher chances of admission: Imperial College or University College London?

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Akash Gaur

Contributor-Level 10

The Imperial College acceptance rate of 25% places it among the most selective higher educational institutes in the region. IC London is only known to accept about 25 applications of every 100 it receives which makes it very competitive to get into.

University College London in comparison is slightly less picky, with an acceptance rate of 30%, UC London is easier to secure admissions at than Imperial College.

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11 months ago

What was the highest package offered at Maya Devi University placements?

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Contributor-Level 10

The highest package offered during the 2024 placement drive at Maya Devi University stood at INR 48 LPA. The same is presented below:

Particulars	Placement Statistics (2024)
the highest package	INR 48 LPA

Note: The placement year is not mentioned on the official website. Hence, the placement data mentioned above is assumed for the year 2024.

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11 months ago

How are the placement opportunities at Maya Devi University?

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Maya Devi University Pharmacy

Contributor-Level 10

Maya Devi University is a well-known institute of offering Management and Technology courses in Uttarakhand. When it comes to the placements, the institute offers good placements annually. The key highlights of Maya Devi University placements for the Class of 2024 are tabulated below:

Particulars	Placement Statistics (2024)
the highest package	INR 48 LPA
Average package	INR 5.48 LPA
Students placed	400+
Total recruiters	500+
Placement rate	92%

Note: The placement year is not mentioned on the official website. Hence, the placement data mentioned above is assumed for the year 2024.

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11 months ago

Does Maya Devi University provide 100% placements for BPharma?

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JIET Group of Institutions Colleges in Jodhpur

Contributor-Level 10

Maya Devi University has a good track record of placements for its BPharma course. The course-wise placement stats are yet to be released, However, the institute recorded 92% placements for the overall Class of 2024. The key highlights of Maya Devi University placements for the Class of 2024 are tabulated below:

Particulars	Placement Statistics (2024)
the highest package	INR 48 LPA
Average package	INR 5.48 LPA
Students placed	400+
Total recruiters	500+
Placement rate	92%

Note: The placement year is not mentioned on the official website. Hence, the placement data mentioned above is assumed for the year 2024.

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11 months ago

What is the placement package offered during JIET Group of Institutions placements?

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Please provide the latest information on results of Tamil Nadu 10th. How does one get the result without roll number?

Contributor-Level 10

The highest package, average package and the lowest package recorded at JIET Group of Institutions placements is provided in the table below:

Particulars

Placement Statistics (2024)

the highest Package

INR 12.47 LPA (UG)

INR 5.75 LPA (PG)

Average Package

INR 4.29 LPA (UG)

INR 3.36 LPA (PG)

the lowest Package

INR 1.2 LPA (UG)

INR 2.6 LPA (PG)

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11 months ago

Exam Tamilnadu 10th

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11 months ago

Q.3.19 Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10²³).

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Contributor-Level 10

3.19 (a) Alloys of metal usually have greater resistivity than that of their constituent metals.

(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin is nearly independent of increase of temperature.

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 10²².

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11 months ago

Who are the top recruiters of Maya Devi University?

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The following is a list of few top recruiters who were part of the recent Maya Devi University placements:

Maya Devi University Top Recruiters
Google	Amazon	TCS
ITC	Accenture	Fedex
Pfizer	Lupin	Glenmark
Cisco	IBM	Tata Group
Godrej	The Oberoi Group	Deloitte

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11 months ago

What is the average package offered at Maya Devi University?

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