Question

Show how would you join three resistors, each of resistance 9 $\Omega$ so that the equivalent resistance of the combination is (a) 13.5 $\Omega$ (b) 6 $\Omega$ ?

OR

(a) Write Joule's law of heating.

(b) Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.

Answer 1

(a) To get an equivalent resistance of 13.5 $\Omega$, a the resistances should be connected as shown in the figure given below :

So,

$\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}$

= $\frac{1}{9}+\frac{1}{9}$

= $\frac{1+1}{9}=\frac{2}{9}$

$\frac{1}{R_P}=\frac{2}{9}$

$R_P=\frac{9}{2}=4.5\Omega$

$R_S=R_3+4.5\Omega$

= $9\Omega +4.5\Omega$

= $13.5\Omega$

(b) To get an equivalent resistance of 6 $\Omega$, the resistances should be connected as shown in the figure given below :
$R_S=R_1+R_2$

= 9+9

= $18\Omega$

Now both the resistors are in parallel with each other so,

$R_P=\frac{1}{18}+\frac{1}{9}$

= $\frac{1+2}{18}=\frac{3}{18}$

=$\frac{1}{6}\Omega$

So, $R_P=6\Omega$

OR

(a) According to Joule's law of heating, the heat produced in a wire is directly proportional to
(i) square of current $\left(I^2\right)$

(ii) resistance of wire (R),

(iii) time (t) for which current is passed.

Thus, the heat produced in the wire by current in time 't' is
H $\infty I^2$ Rt

or H = $K{I}^2$ Rt

But K=1, H = $I^2$ Rt

(b) We know that, $P=V\times I$

$\Rightarrow I=\frac{P}{V}$

First lamp : Let power of first lamp be $P_1$ = 100 W, V = 220 volt and let $I_1$ be the current produced by first lamp.
$I_1=\frac{P_1}{V}=\frac{100}{220}$ = 0.45 A

Second lamp : Let power of second lamp be $P_2$ = 60 W, let voltage be V = 220 volt. Let $I_2$ be the current produced by second lamp.
$I_2=\frac{P_2}{V}=\frac{60}{220}=0.27$ A

So, Total current = $I_1+I_2$

= 0.45+0.27

=0.72 A