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What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

The advantages of connecting electrical devices in parallel with the battery are:

1. Each device gets the full battery voltage.
2. The parallel circuit divides the current through the electrical devices. Each device gets proper current depending on its resistance.
3. Other devices do not get affected if one device is switched OFF/ON. 

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Compute the heat generated while transferring 96,000 coulomb of charge in one hour through a potential difference of 50 V.

Given,

Charge transferred, Q = 96, 000 C
Time, t = 1 hour = 3600 s
Voltage, V = 50 V
Heat generated, H = V×Q
                            = 50 V x 96,000 C
                            = 48,00,000 J 

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What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

(a) Highest resistance can be obtained by connecting the four coils in series. Then
R = R1 + R2 + R3 +R4
R = 4 + 8 + 12 + 24 = 48 Ω. 

(b) Lowest resistance can be obtained by connecting the four coils in parallel.
Then

1 over straight R space equals space 1 fourth plus 1 over 8 plus 1 over 12 plus 1 over 24 space equals 12 over 24 space equals space 1 half

or      straight R space equals space 2 space straight capital omega. space space space space space Ans. space

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Why does the cord of an electric heater not glow while the heating element does?


Current carried by the cord and the heating element of an electric heater is same. But, due to the high resistance of the heating element it becomes hot and starts glowing.
We have,
                        ( H = I2RT) 

And, the cord remains cold due to its low resistance and does not glow.
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How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?


Given three resistors of resistances 2 Ω, 3 Ω and 6 Ω. 

(a) A total resistance of 4 Ω can be obtained by connecting the 2 Ω resistance in series with the parallel combination of 3 Ω and 6 Ω.

That is,
                   Equivalent resiastance, R = R1+R2R3R2+R3                                                =2+3×63+6                                              = 4Ω.   

(b) We can obtain a total resistance of 1 Ω by connecting resistances of 2 Ω, 3 Ω and 6 Ω  in parallel with each other.

              1R= 1R1+1R2+1R3     =12+13+16      =11R = 1 Ω.  
 
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