question_answer

For the series combination of three resistors establish the relation:

\[\mathbf{R = }{{\mathbf{R}}_{\mathbf{1}}}\mathbf{+ }{{\mathbf{R}}_{\mathbf{2}}}\mathbf{+ }{{\mathbf{R}}_{\mathbf{3}}}\]

where the symbols have their usual meanings.

Calculate the equivalent resistance of the combination of three resistors of \[\mathbf{6}\,\mathbf{\Omega ,}\,\mathbf{9}\,\mathbf{\Omega }\] and \[\mathbf{18}\,\mathbf{\Omega }\] joined in parallel.

Answer:

Same current (I) flows through different resistances, when these are joined in series, as shown in the figure.

Let R be the combined resistance, then

\[V=IR\]

Also,     \[{{V}_{1}}=I{{R}_{1}},\text{ }{{V}_{2}}=I{{R}_{2}},{{V}_{3}}=I{{R}_{3}}\]

\[\because \]       \[V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}\]

\[\therefore \]      \[IR=I{{R}_{1}}+I{{R}_{2}}+I{{R}_{3}}\]

\[\Rightarrow \]   \[IR=I({{R}_{1}}+{{R}_{2}}+{{R}_{3}})\]

\[R={{R}_{1}}+{{R}_{2}}+{{R}_{3}}\]

Now,     \[{{R}_{1}}=6\,\Omega ,\text{ }{{R}_{2}}=\,9\,\Omega ,\text{ }{{R}_{3}}=18\,\Omega \]

In parallel combination,

\[\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}\]

\[\Rightarrow \] \[\frac{1}{R}=\frac{1}{6}+\frac{1}{9}+\frac{1}{18}=\frac{3+2+1}{18}\]

\[=\frac{6}{18}=\frac{1}{3}\]

\[\Rightarrow \]   \[\frac{1}{R}=\frac{1}{3}\]

\[\Rightarrow \]   \[R=3\,\Omega \]