Question

a) Two lamps rated 100w,220v and 10w,220v are connected in parallel to 220 v supply. Calculate the total current passing through the circuit.
B) Two resistors X and Y of resistances 2 ohm and 3 ohm respectively are first joined in parallel and then in series in each case the voltage supplied is 5v.
Calculate the voltage across the 3 ohm resistor in the series combination of resistors

Answer 1

Explanation:

Answer of a) 0.5ampere

B) 3volt

Answer 2

A)Total current passing through the circuit

= 0.5 amp.

B)Voltage across the 3 ohm resistor = 3 volt.

Given:

A) Two lamps rated 100w,220v and 10w,220v are connected in parallel to 220    v supply.

B) Two resistors X and Y of resistances 2 ohm and 3 ohm respectively.

  voltage supplied is 5v.

To find:

A) Total current passing through the circuit.

B )Voltage across the 3 ohm resistor in the series combination of resistors.

Formula used:

Ohms law ⇒ voltage = current×resister

Power = voltage × current

Explanation:

A) Power = voltage × current

For first bulb 100w,220v

    P = VI ⇒ $I_{1}$ =$\frac{P}{V}$

                   $I_{1}$=$\frac{100}{220}$ = $\frac{10}{22}$  .................1

For 2nd bulb 10w,220v

 P = VI ⇒ $I_{2}$ =$\frac{P}{V}$

                   $I_{2}$=$\frac{10}{220}$ = $\frac{1}{22}$   .......................2

Adding 1 and 2

$I_{eq}$ =   $I_{1}$+  $I_{2}$

    =  $\frac{10}{22}$ + $\frac{1}{22}$ = $\frac{10+1}{22}$ = $\frac{11}{22}$

$I_{eq}$ = 0.5 amp.

B) For series combination

$R_{eq}$ =  $R_{1}$+  $R_{2}$

$R_{eq}$ =  2+3

     = 5 ohm.

Voltage = 5v ........ ...........Given

Voltage = current×resister

           5 =current× 5

   current = 1 amp.

Voltage across the 3 ohm resistor.

V=IR

V = 3×1

V=3v

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