UrbanPro
true

Find the best tutors and institutes for Class 12 Tuition

Find Best Class 12 Tuition

Please select a Category.

Please select a Locality.

No matching category found.

No matching Locality found.

Outside India?

Learn Exercise 3 with Free Lessons & Tips

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?

Emf of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Maximum current drawn from the battery = I

According to Ohm’s law,

The maximum current drawn from the given battery is 30 A.

Comments

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Emf of the battery, E = 10 V

Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.

Comments

(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?

(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

(a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the combination is given by the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

(b) Current flowing through the circuit = I

Emf of the battery, E = 12 V

Total resistance of the circuit, = 6 Ω

The relation for current using Ohm’s law is,

Potential drop across 1 Ω resistor = V1

From Ohm’s law, the value of Vcan be obtained as

V1 = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = V2

Again, from Ohm’s law, the value of Vcan be obtained as

V2 = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = V3

Again, from Ohm’s law, the value of Vcan be obtained as

V3 = 2 × 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.

Comments

At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is .

Room temperature, T = 27°C

Resistance of the heating element at TR = 100 Ω

Let T1 is the increased temperature of the filament.

Resistance of the heating element at T1R1 = 117 Ω

Temperature co-efficient of the material of the filament,

Therefore, at 1027°C, the resistance of the element is 117Ω.

Comments

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?

Length of the wire, l =15 m

Area of cross-section of the wire, a = 6.0 × 10−7 m2

Resistance of the material of the wire, R = 5.0 Ω

Resistivity of the material of the wire = ρ

Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10−7 Ω m.

Comments

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

Temperature, T1 = 27.5°C

Resistance of the silver wire at T1R1 = 2.1 Ω

Temperature, T2 = 100°C

Resistance of the silver wire at T2R2 = 2.7 Ω

Temperature coefficient of silver = α

It is related with temperature and resistance as

Therefore, the temperature coefficient of silver is 0.0039°C−1.

Comments

Determine the current in each branch of the network shown in fig 3.30:

Current flowing through various branches of the circuit is represented in the given figure.

I1 = Current flowing through the outer circuit

I2 = Current flowing through branch AB

I3 = Current flowing through branch AD

I2 − I4 = Current flowing through branch BC

I3 + I4 = Current flowing through branch CD

I4 = Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

10I2 + 5I4 − 5I3 = 0

2I2 + I4 −I3 = 0

I3 = 2I2 + I4 … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I2 − I4) − 10(I3 + I4) − 5I= 0

5I2 + 5I4 − 10I3 − 10I− 5I= 0

5I2 − 10I3 − 20I= 0

I2 = 2I3 + 4I… (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I2 + 10I− 5I4

3I2 + 2I− I4 = 2 … (3)

From equations (1) and (2), we obtain

I3 = 2(2I+ 4I4) + I4

I3 = 4I+ 8I4 + I4

− 3I3 = 9I4

− 3I4 = + I… (4)

Putting equation (4) in equation (1), we obtain

I3 = 2II4

− 4I4 = 2I2

I2 = − 2I4… (5)

It is evident from the given figure that,

I1 = II2 … (6)

Putting equation (6) in equation (1), we obtain

3I2 +2(II2) − I4 = 2

5I2 + 2I− I4 = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I4) + 2(− 3 I4) −I4 = 2

− 10I4 − 6I− I4 = 2

17I4 = − 2

Equation (4) reduces to

I3 = − 3(I4)

Therefore, current in branch 

In branch BC = 

In branch CD = 

In branch AD 

In branch BD = 

Total current = 

Comments

(a) In a metre bridge [Fig. 3.27], the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if X and Y are interchanged.

(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

A metre bridge with resistors X and Y is represented in the given figure.

(a) Balance point from end A, l1 = 39.5 cm

Resistance of the resistor Y = 12.5 Ω

Condition for the balance is given as,

Therefore, the resistance of resistor X is 8.2 Ω.

The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.

(b) If and Y are interchanged, then l1 and 100−l1 get interchanged.

The balance point of the bridge will be 100−l1 from A.

100−l= 100 − 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.

(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

Comments

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Emf of the storage battery, E = 8.0 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Effective voltage in the circuit = V1

is connected to the storage battery in series. Hence, it can be written as

V1 = V − E

V= 120 − 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

Voltage across resistor R given by the product, IR = 7 × 15.5 = 108.5 V

DC supply voltage = Terminal voltage of battery + Voltage drop across R

Terminal voltage of battery = 120 − 108.5 = 11.5 V

A series resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.

Comments

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?

Emf of the cell, E1 = 1.25 V

Balance point of the potentiometer, l1= 35 cm

The cell is replaced by another cell of emf E2.

New balance point of the potentiometer, l2 = 63 cm

Therefore, emf of the second cell is 2.25V.

Comments

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m−3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10−6 m2 and it is carrying a current of 3.0 A.

 
 
 

Number density of free electrons in a copper conductor, n = 8.5 × 1028 m−3 Length of the copper wire, l = 3.0 m

Area of cross-section of the wire, A = 2.0 × 10−6 m2

Current carried by the wire, I = 3.0 A, which is given by the relation,

nAeVd

Where,

e = Electric charge = 1.6 × 10−19 C

Vd = Drift velocity 

Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 × 104 s.

Comments

A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 ×

Supply voltage, V = 230 V

Initial current drawn, I1 = 3.2 A

Initial resistance = R1, which is given by the relation,

Steady state value of the current, I2 = 2.8 A

Resistance at the steady state = R2, which is given as

Temperature co-efficient of nichrome, α = 1.70 × 10−4 °C −1

Initial temperature of nichrome, T1= 27.0°C

Study state temperature reached by nichrome = T2

Tcan be obtained by the relation for α,

Therefore, the steady temperature of the heating element is 867.5°C

Comments

How helpful was it?

How can we Improve it?

Please tell us how it changed your life *

Please enter your feedback

Please enter your question below and we will send it to our tutor communities to answer it *

Please enter your question

Please select your tags

Please select a tag

Name *

Enter a valid name.

Email *

Enter a valid email.

Email or Mobile Number: *

Please enter your email or mobile number

Sorry, this phone number is not verified, Please login with your email Id.

Password: *

Please enter your password

By Signing Up, you agree to our Terms of Use & Privacy Policy

Thanks for your feedback

About UrbanPro

UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.

X

Looking for Class 12 Tuition Classes?

Find best tutors for Class 12 Tuition Classes by posting a requirement.

  • Post a learning requirement
  • Get customized responses
  • Compare and select the best

Looking for Class 12 Tuition Classes?

Find best Class 12 Tuition Classes in your locality on UrbanPro

Post your learning requirement

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more