(a) Two lamps rated 100 W, 220 V, and 10 W, 220 V are connected in parallel to 220 V supply. Calculate the total current through the circuit.(b) Two resistors X and Y of resistances 2Ω and 3Ω respectively are first joined in parallel and then in series. In each case, the voltage supplied is 5 V.(i) Draw circuit diagrams to show the combination of resistors in each case.(ii) Calculate the voltage across the 3Ω resistor in the series combination of resistors.

 The resistance of Y = $R_2$ = $3\Omega$

Voltage, $V$ = 6V

To find: Voltage across the $3\Omega$ resistor in the series combination of resistors.

Solution:

We know that equivalent resistance (or, total resistance) in a series combination is given as:

$R_{eq}=R_1+R_2$

Putting the value of $R_1\ and\ R_2$ we get-

$R_{eq}=2+3$

$R_{eq}=5\Omega$

We know that current is given as-

$I=\frac {V}{R}$

Here, 

$I=\frac {V}{R_{eq}}$

So, putting the given values we get-

$I=\frac {5}{5}$

$I=1A$

Thus, the currenr through the ciruit is 1A.

Now,

To find the voltage across $3\Omega$ resistors we use the formula of voltage, which is given as-

$V=I\times {R}$

Putting the required values, we get-

$V=1\times {3}$

$V=3V$

Thus the voltage across $3\Omega$ resistor is 3 Volts.