**09 Current Electricity**

**Multiple Choice Questions**

**1. In order to convert a moving coil galvanometer into a voltmeter ………………..**

(A) a high resistance is connected in parallel with the galvanometer

(B) a high resistance is connected in series with the galvanometer

(C) a low resistance is connected in parallel with the galvanometer

(D) a low resistance is connected in series with the galvanometer.

**Ans.** (B) a high resistance is connected in series with the galvanometer

**2. Kirchhoff’s first law is the consequence of Law of Conservation of**(A) mass

(B) energy

(C) charge

(D) momentum

**Ans. **(C) charge

**3. A moving coil galvanometer of resistance ‘ **

**‘ gives full scale deflection for certain current. The shunt resistance required to convert it to measure a current ‘**

**‘ times of initial current is………………..**

(A)

(B)

(C)

(D)

**Ans.** (B)

**4. The resistance of galvanometer is ****. If **** is the resistance used to convert the galvanometer into an ammeter, then the effective resistance of the ammeter is**

(A)

(B)

(C)

(D)

**Ans.** (D)

**5. Kirchhoff’s junction law is equivalent to………………..**

(A) conservation of energy

(B) conservation of charge

(C) conservation of electric potential

(D) conservation of electric flux

**Ans.** (B) conservation of charge

**6. Accuracy of potentiometer can be easily increased by………………..**

(A) increasing resistance of wire

(B) decreasing resistance of wire

(C) increasing the length of wire

(D) decreasing the length of wire

**Ans.** (C) increasing the length of wire

**7. An ideal voltmeter has………………..**

(A) low resistance

(B) high resistance

(C) infinite resistance

(D) zero resistance

**Ans.** (C) infinite resistance

**8. Kirchhoff’s voltage law and current law are respectively in accordance with the conservation of………………..**

(A) charge and momentum

(B) charge and energy

(C) energy and charge

(D) energy and momentum

**Ans.** (C) energy and charge

**9. Instrument which can measure terminal potential difference as well as electro motive force (e.m.f.) is**

(A) Wheatstone’s metrebridge.

(B) Voltmeter.

(C) Potentiometer.

(D) Galvanometer.

**Ans. **(C) Potentiometer.

**10. In potentiometer experiment, if **** is the balancing length for e.m.f. of cell of internal resistance **** and **** is the balancing length for its terminal potential difference when shunted with resistance **** then:**

(A)

(B)

(C)

(D)

**Ans. **(A)

**11. The fraction of the total current passing through the galvanometer is………………..**

(A)

(B)

(C)

(D)

**Ans. **(A)

**12. When unknown resistance is determined by metrebridge, the error due to contact resistance is minimized………………..**

(A) by connecting both the resistances only in one gap.

(B) by interchanging the positions of known and unknown resistance.

(C) by using uniform wire.

(D) by obtaining the null point near the ends of the wire.

**Ans.** (B) by interchanging the positions of known and unknown resistance.

**13. S. I. unit of potential gradient is………………..**(A)

(B)

(C)

(D)

**Ans. **(D)

**14.An ideal ammeter has………………..**

(A) low resistance

(B) high resistance

(C) infinite resistance

(D) zero resistance

**Ans. **(D) zero resistance

**15. If the length of a potentiometer wire is increased by keeping constant potential difference across the wire, then………………..**

(A) null point is obtained at larger distance

(B) there is no change in the null point

(C) potential gradient is increased

(D) null point is obtained at shorter distance

**Ans. **(A) null point is obtained at larger distance

**16. In potentiometer experiment, the cell balances at a length of ****. When the cell is shunted by a resistance of ****, the balancing length becomes ****. The internal resistance of the cell is………………..**

(A)

(B)

(C)

(D)

**Ans. **(B)

**Theory Questions**

**9.2 Kirchhoff’s Laws of Electrical Network**

**State Kirchhoff’s laws of electrical network.**

**OR**

**State Kirchhoff’s laws of electricity.**

**Ans:**

**i. Kirchhoff’s first law (Current law or junction law):**

Statement: The algebraic sum of the currents at a junction is zero in an electrical network.

i.e.,

where is the current in the conductor at a junction having conductors.

ii. Kirchhoff’s second law (Voltage law or loop theorem):

Statement: The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emfs) in a closed loop is zero.

i.e.,

**9.3 Wheatstone Bridge**

**Obtain the balancing condition of Wheatstone’s network.**

**Ans: Construction:**

i. Four resistances and are connected to form a quadrilateral as shown in the figure. ii. A battery of emf along with a key is connected between the points and such that point is at higher potential with respect to the point .

iii. A galvanometer of internal resistance is connected between points and .

Working:

i. When the key is closed, current I flow through the circuit. It divides into and at point . is the current through and is the current through S.

ii. The current gets divided at point . Let be the current flowing through the galvanometer. The currents flowing through and are and respectively.

iii. Consider the loop ABDA, applying Kirchhoff’s voltage law in the clockwise sense shown in the loop.

iv. Now consider loop BCDB, applying Kirchhoff’s voltage law in the clockwise sense as shown in the loop,

v. The bridge is said to be balanced, when the current passing through the galvanometer is zero. i.e., . This condition can be obtained by adjusting the values of and .

vi. Substituting in equation (1),

Substituting in equation (2),

Dividing equation (3) by equation (4),

This is the balancing condition for Wheatstone bridge.

**Draw a neat labelled circuit diagram to determine resistance of a galvanometer using Kelvin’s method.**

**Ans:**

**State any ‘two’ possible sources of errors in metrebridge experiment. How can they be minimised?**

**Ans: Sources of errors:**

i. The cross section of the wire may not be uniform.

ii. The ends of the wire are soldered to the metallic strip where contact resistance is developed, which is not taken into account.

iii. The measurements of and may not be accurate.

**To minimize the errors**

i. The value of is so adjusted that the null point is obtained around middle one third of the wire (between and ) so that percentage error in the measurement of and are minimum and nearly the same.

ii. The experiment is repeated by interchanging the positions of unknown resistance and known resistance box R.

iii. The jockey should be tapped on the wire and not slide. The jockey is used to detect whether there is a current through the central branch. This is possible only by tapping the jockey.

[Any two sources of errors and their minimization]

**Describe Kelvin’s method to determine the resistance of a galvanometer by using a metrebridge.**

**OR**

**Describe meterbridge experiment to determine resistance of galvanometer by Kelvin.**

**Ans:** Refer Subtopic 9.3: Q. No. 2 for diagram Working:

i. The galvanometer whose resistance is to be determined is connected in one gap and a known resistance in the other gap.

ii A suitable resistance is taken in the resistance box. The current is sent round the circuit by closing the key. Without touching the jockey at any point of the wire, the deflection in the galvanometer is observed. iii. The rheostat is adjusted to get a suitable deflection around of range.

iv. Now, the jockey is tapped at different points of the wire and a point of contact for which, the galvanometer shows no change in the deflection, is found.

v. As the galvanometer shows the same deflection with or without contact between the point and , these two points must be equipotential points.

vi. The length of the bridge wire between the point and the left end of the wire is measured. Let be the length of the segment of wire opposite to the galvanometer and be the length of the segment opposite to the resistance box.

**Calculation:**

i. Let and be the resistance of the two parts of the wire and respectively.

ii. Since bridge is balanced,

But,

where, is length of wire

Using this formula, the unknown resistance of the galvanometer can be calculated.

**Explain with neat circuit diagram, how will you determine the unknown resistance ‘****‘ by using a metrebridge experiment.**

**[Mar 18]**

**Ans: Construction:**

i. Metrebridge consists of a one metre long wire of uniform cross section, stretched on a metre scale which is fixed on a wooden table.

ii. The ends of the wire are fixed below two L shaped metallic strips. A single metallic strip separates the two L shaped strips leaving two gaps, left gap and right gap.

iii. Usually, an unknown resistance is connected in the left gap and a resistance box is connected in the other gap.

iv. One terminal of a galvanometer is connected to the point on the central strip, while the other terminal of the galvanometer carries the jockey (J). Temporary contact with the wire can be established with the help of the jockey.

v. A cell of emf along with a key and a rheostat are connected between the points and .

**Working:**

i. A suitable resistance is selected from resistance box.

ii. The jockey is brought in contact with at various points on the wire and the balance point (null point), is obtained. The galvanometer shows no deflection when the jockey is at the balance point (point D).

iii. Let the respective lengths of the wire between and , and that between and be and .

iv. Then using the balancing conditions,

where and are resistance of the parts and of the wire respectively.

v. If and are lengths of the parts and of the wire is its specific resistance of the wire, and is its area of cross section of wire then

From equations (1), (2) and (3),

Thus, knowing and , the value of the unknown resistance can be determined.

**State Kirchhoff’s second law for a closed circuit. How will you minimize the errors in Kelvin’s method?**

**Ans:**

i. Kirchhoff’s second law for a closed circuit:

Statement: The algebraic sum of the potential differences (products of current and resistance) and the electromotive forces (emfs) in a closed loop is zero.

i.e., ii. Ways to minimize the errors:

a. The value of is so adjusted that the null point is obtained around middle one third of the wire (between and ) so that percentage error in the measurement of and are minimum and nearly the same.

b. The experiment is repeated by interchanging the positions of unknown resistance and known resistance box R.

c. The jockey should be tapped on the wire and not slided. The jockey is used to detect whether there is a current through the central branch. This is possible only by tapping the jockey.

**9.4 Potentiometer**

**Describe how a potentiometer is used to compare the emfs of two cells by combination method.**

**Ans:**

i. The emfs of two cells can be compared using sum and difference method.

ii. When two cells are connected such that the positive terminal of the first cell is connected to the negative terminal of the second cell, the emf of the two cells are added up and the effective emf of the combination of two cells is . This method of connecting two cells is called the sum method.

iii. When two cells are connected such that their negative terminals are together or their positive terminals are connected together, then their emf oppose each other and effective emf of the combination of two cells is (Considering ). This method of connecting two cells is called the difference method.

iv. Circuit for sum and difference method is connected as shown in below figure. When keys and are closed the cells and are in the sum mode. The null point is obtained using the jockey.

v. Let be the length of the wire between the null point and the point . This corresponds to the .

where is the potential gradient along the potentiometer wire.

vi. Now the key and are kept open and keys and are closed. In this case the two cells are in the difference mode.

vii. Again the null point is obtained. Let be the length of the wire between the null point and the point . This corresponds to .

viii. Dividing equation (1) by equation (2),

By componendo and dividendo method,

Thus, emf of two cells can be compared using sum and difference method.

**Describe how a potentiometer is used to compare the e.m.f.s of two cells by connecting them separately.**

**Ans:**

i. A potentiometer circuit is set up by connecting a battery of emf , with a key and a rheostat such that point is at higher potential than point .

ii. The cells whose emfs are to be compared are connected with their positive terminals at point and negative terminals to the extreme terminals of a two way key and .

ii. The central terminal of the two ways key is connected to a galvanometer. The other end of the galvanometer is connected to a jockey .

iv. Key is closed and then, key is closed and key is kept open. Therefore, the cell of emf comes into circuit.

v. The null point is obtained by touching the jockey at various points on the potentiometer wire .

vi. Let be the length of the wire between the null point and the point . Here, corresponds to emf of the cell. Therefore,

where is the potential gradient along the potentiometer wire.

vii. Now key is kept open and key is closed. The cell of emf now comes in the circuit. Again, the null point is obtained with the help of the Jockey.

viii. Let be the length of the wire between the null point and the point A. Here corresponds to the emf of the cell.

ix. Dividing equation (1) by equation (2),

Thus, emfs of the two cells can be compared and if any one of the emfs is known, the other can be determined.

**State and explain principle of potentiometer.**

**[Oct 13]**

**Ans: Principle:**

The potential difference between two points on the wire is directly proportional to the length of the wire between them provided the wire is of uniform cross section, the current through the wire is the same and temperature of the wire remains constant.

**Explanation:**

i. A potentiometer consists of a long wire of length and resistance having uniform cross sectional area .

ii. A cell of emf having internal resistance is connected across as shown in the figure.

iii. When the circuit is switched on, current I pass through the wire.

Current through

iv. Potential difference across ,

….[From equation (1)]

v. Therefore, the potential difference per unit length of the wire is,

As long as remains constant, will remain constant.

vi. is known as potential gradient along and is denoted by K. Potential gradient can be defined as potential difference per unit length of wire.

vii. Consider a point on the wire at distance from the point , as shown in the figure. If the potential difference between and is , then, i.e.,

viii. Thus, the potential difference between two points on the wire is directly proportional to the length of the wire between the two points, provided (a) the wire is of uniform cross section, (b) the current through the wire is the same and (c) temperature of the wire remains constant.

**State the advantages of potentiometer over voltmeter.**

**Ans:**

i. Potentiometer is more sensitive than a voltmeter.

ii. A potentiometer can be used to measure a potential difference as well as an emf of a cell. A voltmeter always measures terminal potential difference, and as it draws some current, it cannot be used to measure an emf of a cell.

iii. Measurement of potential difference or emf is very accurate in the case of a potentiometer. A very small potential difference of the order volt can be measured with it. Least count of a potentiometer is much better compared to that of a voltmeter.

Due to all these reasons potentiometer is preferred over a voltmeter for measuring emf.

**Define potential gradient of the potentiometer wire.****Ans:**Potential gradient is defined as potential difference per unit length of wire.

**Distinguish between potentiometer and voltmeter.**

**Ans:**

0. | meter | Voltmeter |

i. | Its resistance isinfinite. | Its resistance is highbut finite. |

ii. | It does not draw anycurrent from thesource of known e.m.f. | It draws some currentfrom the source ofe.m.f. |

iii. | The potentialdifference measuredby it is equal to actualpotential difference(p.d.). | The potentialdifference measuredby it is less than theactual potentialdifference (p.d.). |

iv. | It has high sensitivity. | It has low sensitivity. |

v. | s e.m.f as | sures only p.d. |

vi. | It is usedinternal rea cell. | ot be used tothe internalce of a cell. |

vii. | It is mor | It is less accurate. |

viii. | give direct | It givesreading. |

ix. | It is not portable. | It is portable. |

. | It is used to measurelower voltage valuesonly. | It is used to measurelower as well ashigher voltage values. |

[Any four differences]

**State the formula giving relation between electric field intensity and potential gradient.**

**Ans:** Formula:

Where, is the potential gradient and is the electric field.

**Explain the use of potentiometer to determine internal resistance of a cell.**

**Ans:**

i. The experimental set up for this method consists of a potentiometer wire connected in series with a cell of emf , the key , and rheostat as shown in figure.

iii. A resistance box is connected across the cell through the key . The key is closed and is open.

iv. The circuit now consists of the cell , cell , and the potentiometer wire. The null point is then obtained.

v. Let be length of the potentiometer wire between the null point and the point . This length corresponds to emf .

where is potential gradient of the potentiometer wire which is constant.

vi. Now both the keys and are closed so that the circuit consists of the cell , the cell , the resistance box, the galvanometer and the jockey. Some resistance R is selected from the resistance box and null point is obtained.

vii. The length of the wire between the null point and point is measured. This corresponds to the voltage between the null point and point .

Dividing equation (1) by equation (2),

viii. Consider the loop PQSTP,

and

….[From equation (3)]

Ans: Moving coil galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer, which effectively reduces the resistance of the galvanometer. This low resistance connected in parallel is called as shunt (S).

**What is the value of resistance for an ideal voltmeter?**

**Ans:** For an ideal voltmeter the value of resistance is infinite.

**How will you convert a moving coil galvanometer into an ammeter?**

**Ans:**

The above equation is used to determine the internal resistance of the cell.

**9.5 Galvanometer**

**1 Eryhin how moving coil galvanometer is coverted into a voltmeter. Derive the necewary formula. **

**OR**

**Explain how a moving coil galvanometer can be converted into voltmeter.**

**Ans:**

i. To use a M.C.G as a voltmeter, a high resistance is connected in series with the M.C.G.

ii. A very high resistance is connected in series with the galvanometer for this purpose as shown in figure.

i. To use a M.C.G as an ammeter, a shunt (low resistance) is connected in parallel to the coil of M.C.G.

ii. In the arrangement as shown in the figure, I and are the current through the circuit and galvanometer respectively.

Therefore, the current through shunt is,

iii. Since S and G are parallel, potential difference across them is same.

Equation (1) is useful to calculate the range of current that the galvanometer can measure.

iv. If the current is times current , then . Using this in equation (1),

This is the required shunt to increase the range times.

v. If is the current through the shunt resistance, then the remaining current (I ) will flow through galvanometer.

This equation gives the fraction of the total current through the shunt resistance.

**Numericals**

**9.2 Kirchhoff’s Laws of Electrical Network**

**A voltmeter has a resistance of****. What will be its reading when it is connected across a cell of e.m.f.****and internal resistance****? [July 16]**

**Solution:**

Given:

To find: Reading of voltmeter

Formula:

Calculation: Current through the circuit is given by

From formula,

Ans: The reading on the voltmeter is .

**Find the total current ‘I’ flowing through the following circuit:**

**[July 19]**

**Solution:**

From figure, two resistors are connected in parallel.

. Total resistance in that branch is,

this combination is connected in series with .

Ans: The total current ‘ ‘ flowing through the circuit is 1 A.

**Two cells of emf****and****having respective internal resistance of****and****are connected in parallel, so as to send current in the same direction through an external resistance of****. Find the current through the external resistance. [Mar 22]**

**Solution:**

Given:

Applying Kirchhoff’s law in Loop 1 (ABCDEFGHA),

Applying Kirchhoff’s law in Loop 2 (FGHAKJF),

Substituting equation (ii) in equation (i),

Current through external resistance ‘ ‘

Ans: Current through external resistance is .

**9.3 Wheatstone Bridge**

**Four resistances****and****are connected in a series so as to form Wheatstone’s network. If the network is balanced, find the value of ‘****‘.**

**Solution:**

Given:

To find: Unknown resistance (X).

Formula:

Calculation: From formula,

Ame The unknown resistance is .

**2. Two diametrically opposite points of a metal ring are connected to two terminals of the left gap of metrebridge. The resistance of **** is connected in right gap. If null point is obtained at a distance of **** from the left end, find the resistance of metal ring.**

**Solution:**

Given: ,

To find: Resistance of metal ring

Formula:

Calculation: Resistance of each half segment of the metal ring and these half segments are connected in parallel in the left gap.

From the formula,

Ans: The resistance of the metal ring is .

**An unknown resistance is placed in the left gap and resistance of****is placed in the right gap of a metrebridge. The null point is obtained at****from the left end. Determine the unknown resistance.**

**Solution:**

Given: ,

To find: Unknown resistance (X)

Formula:

Calculation: From formula,

Ans: The value of unknown resistance is .

**Two resistances****and****in the two gaps of a metrebridge gives a null point dividing the wire in the ratio****. If each resistance is increased by****, the null point divides the wire in the ratio****, calculate the value of****and****.**

**Solution:**

Given:

To find: Resistance ( and )

Formula:

Calculation: From condition,

From condition,

[From equation (i)]

From equations (i) and (ii), we have,

Ans: i. The value of resistance is .

ii. The value of resistance is .

**9.4 Potentiometer**

**The potentiometer wire has length****and resistance****. If the current flowing through it is****, what are the balancing lengths when two cells of****. f****and 1.1****are connected so as to**

**i. assist and**

**ii. oppose each other?**

**Solution:**

Given: ,

To find: i. Null point distance when two cells assist each other

ii. Null point distance when two cells oppose each other

Formulae: i.

ii.

**Calculation:**

i. Since,

Here,

When and are connected so as to assist then,

From formula (i),

ii. From formula (ii),

Ans: i. The null point distance when two cells assist each other is .

ii. The null point distance when two cells oppose each other is .

**A cell balances against a length of****on a potentiometer wire when it is shunted by a resistance of****The balancing length reduces by****when it is shunted by a resistance of****Calculate the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.**

**Solution:**

Given: ,

To find: Balancing length , Internal resistance of the cell

Formula:

Calculation: From first condition,

From second condition,

From equation (i) and (ii),

Using equation (i)

Internal resistance,

antilog

Ans: i. The balancing length is .

ii. The internal resistance of the cell is .

**A potentiometer wire has a length of****and a resistance of****. What resistance should be connected in series with a potentiometer wire and a cell of e.m.f.****having internal resistance****to get a potential gradient of****?**

**Solution:**

Given:

,

To find: Series resistance (X)

Formula:

Calculation: From formula,

Ans: A resistance of should be connected in series.

**Resistance of a potentiometer wire is****. A cell of e.m.f.****is balanced at****on this potentiometer wire. Calculate the current and balancing length for another cell of e.m.f. 1.4 V on the same potentiometer wire.**

**Solution:**

Given:

To find: Current (I), balancing length

Formulae:

i.

Calculation: Using formula (i),

Using formula (ii),

Ans: . The flow of current is .

ii. The balancing length for second cell is .

**A potentiometer wire has a resistance per unit length of****. A cell of e.m.f.****balances against****length of the wire. Find the current through potentiometer wire.**

**Solution:**

Given: ,

To find: Current through potentiometer (I).

Formulae: i. ii.

Calculation: From formula (i),

From formula (ii),

Ans: Current through the potentiometer wire is .

**A potentiometer wire has a length of****and resistance of****. It is connected in series with resistance****and a cell of e.m.f****. Calculate the potential gradient along the wire.**

**Solution:**

Given: ,

To find: Potential gradient of wire

Formula:

Calculation: Since,

From formula,

Ans: The potential gradient of wire is .

**When a resistance of****is connected across a cell, its terminal potential difference is balanced by****length of potentiometer wire. When the resistance of****is connected across the same cell, the balancing length is****. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.**

**Solution:**

Given:

,

To find: Balancing length , Internal resistance

Formula:

Calculation: From first condition,

From second condition,

From equation (i) and (ii),

Using equation (i) we get

Internal resistance,

Ans: i. The balancing length is .

ii. The internal resistance of the cell is .

**In a potentiometer the balancing length of the wire is found to be****for a cell of e.m.f. 1.5 V. Find the balancing length of the wire for another cell of e.m.f.****on the same potentiometer.**

**Solution:**

[July 18]

Given:

To find: Balancing length of the wire

Formula:

Calculation: From formula,

Ans: The balancing length of the wire for another cell is .

**When a resistor of****is connected across the cell, its terminal potential difference is balanced by****of potentiometer wire and when a resistance of****is connected across the cell, the terminal potential difference is balanced by****of the same potentiometer wire. Find the balancing length when the cell is in open circuit and the internal resistance of the cell. [Mar 19]**

**Solution:**

Given:

,

To find: Balancing length , Internal resistance

Formula:

Calculation: From first condition,

From second condition,

From equation (i) and (ii),

Using equation (i) we get

Internal resistance,

Ans: i. The balancing length is .

ii. The internal resistance of the cell is .

**When two cells of emfs****and****are connected in series so as to assist each other, their balancing length on potentiometer wire is found to be****. When two cells are connected in series so as to oppose each other, the balancing length is found to be****. Compare the emfs of two cells.**

**Solution:**

Given:

To find: Ratio of emfs of two cells

Formula:

Calculation: From formula,

Ans: The ratio of emfs of two cells is .

**9.5 Galvanometer**

**A voltmeter of resistance****can measure a maximum voltage of 5 volt. How can it be made to measure a maximum voltage of 100 volt?**

**Solution:**

Given:

To Find: High resistance (R)

Formula:

Calculation: We know that,

Now, using formula we get,

Ans: Given voltmeter can be made to measure a maximum voltage of by connecting a resistance of in series with it.

**A galvanometer has a resistance of****It shows full scale deflection when a current of****is passed through it. The only shunt resistance available is****which is not appropriate to convert a galvanometer into an ammeter. How much resistance should be connected in series with the coil of galvanometer so that the range of ammeter is 8 A?**

**Solution:**

Let ‘ ‘ be the resistance connected in series with galvanometer. Here is not sufficient for .

From the figure,

Ans: A resistance of should be connected in series with coil of galvanometer.

**3. The combined resistance of a galvanometer of resistance **** and its shunt is ****. Calculate the value of shunt.**

**Solution:**

Given:

To find: Shunt resistance (S)

Formula:

Calculation: From formula,

Ans: The value of shunt resistance is .

**A moving coil galvanometer has a resistance of****and gives a full scale deflection for a current of****. How will you convert it into a voltmeter having range****?**

**Solution:**

Given: ,

To find: Resistance

Formula:

Calculation: From formula,

Ans: A resistance of should be connected in series.