A potentiometer wire PQ of 1 m length is connected to a standard cell E 1 . Another cell E 2 of emf 1 . 02 V is connected as shown in the circuit diagram, with a resistor R and a switch S . With switch S open, null point is obtained at a distance of 51 cm from P . Calculate iii when switch S is closed will null point shift towards P or Q . Justify your answer.

When switch S is closed no current is drawn from the cell E 2 There is no shift point depends on open the potential gradient along the potentiometer wire. Which depends upon the emf of battery E 1 And resistance of potentiometer wire circuit and length of potentiometer and emf of the cell E 2 Which does not change when switch S is closed

A potentiometer wire PQ of 1 m length is connected to a standard cell E 1 . Another cell E 2 of emf 1 . 02 V is connected as shown in the circuit diagram, with a resistor R and a switch S . With switch S open, null point is obtained at a distance of 51 cm from P . Calculate ii the emf of the cell E 1

The emf of a cell is the sum of the electric potential differences (PDs) produced by a separation of charges (electrons or ions) that can occur at each phase boundary (or interface) in the cell. Expression for emf of the cell EMF EMF = kl PQ = 0 .02 V cm − 1 × 100 cm = 2 V .

A potentiometer wire PQ of 1 m length is connected to a standard cell E 1 . Another cell E 2 of emf 1 . 02 V is connected as shown in the circuit diagram, with a resistor R and a switch S . With switch S open, null point is obtained at a distance of 51 cm from P . Calculate i the potential gradient of the potentiometer wire

Potential Gradient is the decrease in potential per unit length. It is calculated as V / L , where V is the potential difference between two points and L is the distance between two points. The longer the wire the lesser the potential gradient and the greater the sensitivity of the potentiometer. Length of potentiometer wire L = 1 m , Emf E 1 = e , Emf of cell E 2 = 1 . 02 v s - 1 Null point = 51 cm = 0 . 51 m Potential gradient of the wire V = 1 . 02 0 . 51 = 2 V m - 1

A cell of internal resistance r passes current for some time in a resistor R 1 , in another resistance R 2 current is passed for the same time. If same heat is produced in the two resistors them show that r = R 1 R 2 .

As heat produced in similar time is same for two cases, Power dissipated should be same in R 1 and R 2 . Let internal resistance of battery be r and Emf be E . Internal resistance is r , For first resistance R 1 , Heat produced in the resistor, H = I 2 RT ∴ I = E R 1 + r Thus, H = E R 1 + r 2 R 1 t . . . . . . . . . . . . ( i ) And for second resistance R 2 H ' = E R 2 + r 2 R 2 t . . . . . ( ii ) According to the question, H = H ' E R 1 + r 2 R 1 t = E R 2 + r 2 R 2 t On solving we get, R 1 R 2 ( R 2 − R 1 ) = r 2 ( R 2 − R 1 ) ∴ R 1 R 2 = r 2 OR r = R 1 R 2 Hence, proved.

Find out the reading of voltmeter with the help of the given electrical circuit.

Using Kirchhoff's Voltage Law in the circuit - 2 I + 18 - 12 - I = 0 - 3 I + 6 = 0 ∴ I = 2 A V B - 2 × 2 + 18 = V A V A - V B = 18 - 4 V AB = 14 V

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 13

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IMPORTANT

Earn 100

The sides of a cube of the same material are $l$ and $2 l$ respectively. The ratio of their resistances will be

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Important Questions on Current Electricity

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 14

Electromotive force of a cell is

2

volt. What is the meaning of this statement ?

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 15

A cell of

e . m . f . 10 V

and internal resistance

3 Ω

is connected to a resistor in series. If the current in the circuit is

0.5 A

, then find out the resistance of the resistor and terminal voltage across the cell.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 16

If dissipated heat in $9 Ω$ resistor is $36$ watt, then find the heat dissipated in $6 Ω$ resistor and potential difference across the ends of $2 Ω$ resistor.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 17

The length of a wire becomes three times by stretching it. The wire is connected with a source of constant voltage. What would be the effect on

$(i)$ resistance

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 18

The length of a wire becomes three times by stretching it. The wire is connected with a source of constant voltage. What would be the effect on

$(ii)$ resistivity of the material

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 19

The length of a wire becomes three times by stretching it. The wire is connected with a source of constant voltage. What would be the effect on

$(iii)$ drift speed of the free electrons of the wire?

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 3 - Current Electricity>Uttar Pradesh Board-2020>Q 20

What is the principle of a potentiometer ? The resistance of a potentiometer wire is $8 Ω$ . Find the potential difference across the $4 Ω$ resistor, when the sliding jockey is just in contact at the mid-point $C$ of the potentiometer wire :

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