question_answer

Show how would you join three resistors, each of resistance \[\mathbf{9}\,\mathbf{\Omega }\] so that the equivalent resistance of the combination is (a) \[\mathbf{13}\mathbf{.5}\,\mathbf{\Omega }\] (b) \[\mathbf{6}\,\mathbf{\Omega }\]?

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:

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

So,        \[\frac{1}{{{R}_{P}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\]

\[=\frac{1}{9}+\frac{1}{9}\]

\[=\frac{1+1}{9}=\frac{2}{9}\]

\[\frac{1}{{{R}_{P}}}=\frac{2}{9}\]

\[{{R}_{P}}=\frac{9}{2}=4.5\Omega \]

Now,     \[{{R}_{S}}={{R}_{3}}+4.5\Omega \]

\[=9\,\Omega +4.5\,\Omega \]

\[=13.5\,\Omega \]

(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 \[({{I}^{2}}),\]

(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

\[H\,\propto \,{{I}^{2}}Rt\]

or           \[H=K\,{{I}^{2}}Rt\]

But        \[K=1,\] \[H={{I}^{2}}Rt\]

(b) We know that,          P = VI

\[\Rightarrow \]            \[I=\frac{P}{V}\]

First lamp: \[{{P}_{1}}=100\text{ }W,\text{ }V=220\text{ }volt\]

\[{{I}_{1}}=\frac{{{P}_{1}}}{V}=\frac{100}{220}=0.45A\]

Second lamp: \[{{P}_{2}}=60\text{ }W,\text{ }V=220\text{ }volt\]

\[{{I}_{2}}=\frac{{{P}_{2}}}{V}=\frac{60}{220}=0.27A\]

So,    Total current \[={{I}_{1}}+{{I}_{2}}\]

= 0.45 + 0.27

= 0.72 A