Search any question & find its solution
Question:
Answered & Verified by Expert
(a) Three resistors of $2 \Omega, 4 \Omega$ and $5 \Omega$ are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf $20 \mathrm{~V}$ and negligible internal resistance then determine the current through each resistor, and the total current drawn from the battery.
(b) If the combination is connected to a battery of emf $20 \mathrm{~V}$ and negligible internal resistance then determine the current through each resistor, and the total current drawn from the battery.
Solution:
1633 Upvotes
Verified Answer
(a) Given, $\mathrm{R}_1=2 \Omega, \mathrm{R}_2=4 \Omega, \mathrm{R}_3=5 \Omega$, $\mathrm{R}_{\mathrm{p}}=$ ?
(b) $\mathrm{V}=20 \mathrm{~V}, \mathrm{I}_1=?, \mathrm{I}_2=?, \mathrm{I}_3=?, \mathrm{I}=$ ?
(a) By formula, $\frac{1}{\mathrm{R}_p}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}$ $=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}$ $\mathrm{R}_{\mathrm{p}}=\frac{20}{19}$ ohm.
(b) By formula, $I=\frac{\mathrm{V}}{\mathrm{R}}$;
$$
\mathrm{I}_1=\frac{\mathrm{V}}{\mathrm{R}_1}=\frac{20}{2}=10 \mathrm{~A} \text {; }
$$
$$
\begin{aligned}
&\mathrm{I}_2=\frac{\mathrm{V}}{\mathrm{R}_2}=\frac{20}{40}=5 \mathrm{~A} ; \\
&\mathrm{I}_3=\frac{\mathrm{V}}{\mathrm{R}_3}=\frac{20}{5}=4 \mathrm{~A}
\end{aligned}
$$
By relation, $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$
$$
=10+5+4=19 \mathrm{~A} .
$$
(b) $\mathrm{V}=20 \mathrm{~V}, \mathrm{I}_1=?, \mathrm{I}_2=?, \mathrm{I}_3=?, \mathrm{I}=$ ?
(a) By formula, $\frac{1}{\mathrm{R}_p}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}$ $=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{10+5+4}{20}=\frac{19}{20}$ $\mathrm{R}_{\mathrm{p}}=\frac{20}{19}$ ohm.
(b) By formula, $I=\frac{\mathrm{V}}{\mathrm{R}}$;
$$
\mathrm{I}_1=\frac{\mathrm{V}}{\mathrm{R}_1}=\frac{20}{2}=10 \mathrm{~A} \text {; }
$$
$$
\begin{aligned}
&\mathrm{I}_2=\frac{\mathrm{V}}{\mathrm{R}_2}=\frac{20}{40}=5 \mathrm{~A} ; \\
&\mathrm{I}_3=\frac{\mathrm{V}}{\mathrm{R}_3}=\frac{20}{5}=4 \mathrm{~A}
\end{aligned}
$$
By relation, $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3$
$$
=10+5+4=19 \mathrm{~A} .
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.