Oops! It appears that you have disabled your Javascript. In order for you to see this page as it is meant to appear, we ask that you please re-enable your Javascript!

Â

**Q.1 Â Â Â What does an electric current mean?
**

**Q.2 Â Â Â Define the unit of current.
**

Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.

Therefore, 1 ampere =Â

**Q.3 Â Â Â Calculate the number of electrons constituting one coulomb of charge.
Sol.Â Â Â Â Â **We know that charge over 1 electron = 1.6 Ã— 10

Thus, 1.6 Ã— 10^{â€“19}Â C of charge = 1 electron

Therefore, 1 C of charge = Electrons

= electrons =Â Â electrons = 6.25 Ã— 10^{18} electrons

**Q.4 Â Â Â Name a device that helps to maintain a potential difference across a conductor.
**

**Q.5 Â Â Â What is meant by saying that the potential difference between two points is 1 V?
**

**Q.6 Â Â Â How much energy is given to each coulomb of charge passing through a 6 V battery?**

** Sol.Â Â Â Â Â **Given, Charge Q = 1C, Potential difference, V = 6V

Therefore, Energy i.e. Work done, W =?

We know that, V =Â

Therefore, 6V =Â

W = 6V Ã— 1C = 6J

Thus, required energy = 6J

**Q.7 Â Â Â On what factors does the resistance of a conductor depend?
**

(a) Nature of conductor

(b) Length of conductor

(c) Area of cross section of conductor

**Q.8 Â Â Â Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
**

**Q.9 Â Â Â Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Sol.Â Â Â Â **Since Resistance (R) =Â

Therefore, if potential difference between two ends of the component will be halved, and resistance remains constant, then electric current would also be halved.

**Q.10 Â Â Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
**

**Q.11 Â Â Use the data in Table 12.2 to answer the following â€“Â Â Â Â Â Â Â (a) Which among iron and mercury is a better conductor?**

Â Â Â Â Â Â Â Â (b)Silver

**Q.12 Â Â Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Î© resistor, an 8 Î© resistor, and a 12 Î© resistor, and a plug key, all connected in series.
Sol.**

**Q.13 Â Â Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Î© resistor. What would be the readings in the ammeter and the voltmeter?**

**Sol.**

The total resistance in the circuit = Sum of the resistances of all resistors

= 5 Î© + 8 Î© + 12 Î© = 25 Î©

We know;

R= or, or,Â A

Since, resistances are connected in series, thus electric current remains the same through all resistors.

Here we have,

Electric current, I = 0.24A

Resistance, R = 12Î©

Thus, potential difference, V through the resistor of 12Î© = I x R

Or, V = 0.24A x 12Î© = 2.88 V

Thus, reading of ammeter = 0.24A

Reading of voltmeter through resistor of 12Î© = 2.88V

**Q.14 Â Â Judge the equivalent resistance when the following are connected in parallel â€“ Â Â Â Â Â Â Â (a) 1 Î© and 106 Î©, Â Â Â Â Â Â Â Â Â Â Â (b) 1 Î© and 103 Î©, and 106 Î©.
**

SinceÂ

when resistors are connected in paralle

(a) 1 Î© and 106 Î©

Thus,Â

Thus,Â

Thus, equivalent resistance ofÂ andÂ are connected in parallel =Â

(b)Â **Â **1 Î© and 103 Î©, and 106 Î©

Thus,Â

Thus,Â

Thus, equivalent resistance of **Â **1 Î©, 103 Î© and 106 Î©Â are connected in parallel =Â

**Q.15 Â Â An electric lamp of 100 Î©, a toaster of resistance 50 Î©, and a water filter of resistance 500 Î© are connected in parallel to a 220 V source.
**

Given:

R1 = 100Î©, R2 = 50 Î©, R3 = 500 Î© all are connected in parallel

Potential difference = 220V

Thus,Â

Therefore,Â

Electric current (I) through the circuit =Â

For electric iron

Since it takes as well current as three appliances, thus electric current through it = 7.04A

The electric current = 7.04 A and potential difference = 220 V

Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Î©

Thus, electric current through the electric iron = 7.04A

Resistance of electric iron = 31.25 Î©

**Q.16 Â Â What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
**

Advantages of connecting electrical appliances in parallel instead of connecting in series:

(a) Voltage remains same in all the appliances.

(b) Lower total effective resistance

**Q.17 Â Â How can three resistors of resistances 2 Î©, 3 Î©, and 6 Î© be connected to give a total resistance of (a) 4 Î©, (b) 1 Î©?
Sol.**

(a) When resistors having resistance equal to 3 Î© and 6 Î© are connected in parallel and one having resistance equal to 2 Î© is connected in series

Let total resistance due to resistors having resistance equal to 6 Î© and 3 Î© = R_{1
}Therefore,Â

ThusÂ

Now, total effective resistance in the circuit = R_{1}Â + 2 Î© = 2 Î© + 2 Î© = 4 Î©

Hence, when resistors having resistance equal to 3 Î© and 6 Î© are connected in parallel and one having resistance equal to 2 Î© is connected in series, then the total effective resistance in the circuit = 4 Î©

(b) When all the three resistance is connected in parallel then

Thus,Â

When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Î©

**Q.18 Â Â What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Î©, 8 Î©, 12 Î©, 24 Î©?
**

When all the resistors are connected in series

Then theÂ

Thus, R = 2 Î©

(b) When all the resistors are connected in series

Thus, total resistance = 4 Î© + 8 Î© + 12 Î© + 24 Î© = 48 Î©

Thus, highest resistance = 48 Î©

Lowest resistance = 2 Î©

Page Number â€“ 218

**Q.19 Â Â Why does the cord of an electric heater not glow while the heating element does?
Sol.**

The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.

**Q.20 Â Â Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
**

Given, Potential difference (V) = 50V,

Charge (Q) = 96000 coulomb

Time, t = 1 hour = 60 X 60 s = 3600 s

Heat produced =?

We know that electric current (I) =Â

We know that heat produced (H) in the given time (t) = VIt

Joule

**Q.21 Â Â An electric iron of resistance 20 Î© takes a current of 5 A. Calculate the heat developed in 30 s.
Sol.**

Given, Electric current (I) = 5A, Resistance (R) = 20 Î©, Time (t) = 30s

Therefore, Heat produced (H) =?

Since,Â

We know that, Heat produced (H) = VIt

Answer

**Q.22 Â Â What determines the rate at which energy is delivered by a current?
**

Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.

**Q.23 Â Â An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Sol.Â Â Â Â **Given, Electric current (I) = 5A,

Potential difference (V) = 220V,

Time (t) = 2h = 2 x 60 x 60 s = 7200 s

Power (P) =?

Energy consumed =?

We know that, P = VI = 220V Ã— 5A = 1100W

We know that energy consumed by the electric appliance = P Ã— t

Energy consumed = 1100 W Ã— 7200 s = 7920000 J = 7.92 Ã— 10^{6} JÂ Â

Contact Us

good

Thanks sir

I like this sites

Jabardast, lajawab answers luv u drone

Longer but easy method

Solutions are helpful for my mid term exam .

THANK YOU SIR.....

#...Best for me..#

Am amarendra,

am happy ur method see am read so very very suprrrr

Shandar app

Thanks sir but I want to see the second lecture of life processes in biology

If you want any information regarding this course please call us on 8287971571 or 0261-4890016.

Nice app sir

Thanks sir for clearing aur doubt .

It's really a nice app

Very good

thnku solutions are very helpful

thanks very much easy

sooo much easy to understand.

thankyou very much

Lengthy but easy to understand thanks

Thanks sir

very nice sir

I like this sites

I like this site very much

Very easy to understand. It is very interesting to read.