Question 14 Marks
If two or more resistances are connected in such a way that the same potential difference gets applied to each of them, then they are said to be connected in parallel.
The current flowing through the two resistances in parallel is, however, not the same. When we have two or more resistances joined in parallel to one another, then the same current gets additional paths to flow and the overall resistance decreases.
i. Three resistances, $2 \Omega, 6 \Omega$ and $8 \Omega$ are connected in parallel, then what will be the equivalent resistance?
ii. A wire of resistance $12 \Omega$ is cut into three equal pieces and then twisted their ends together, then what will be the equivalent resistance?
iii. Three resistances are connected as shown. Calculate the equivalent resistance between $A$ and $B$
OR
Find the current in each resistance.
The current flowing through the two resistances in parallel is, however, not the same. When we have two or more resistances joined in parallel to one another, then the same current gets additional paths to flow and the overall resistance decreases.
i. Three resistances, $2 \Omega, 6 \Omega$ and $8 \Omega$ are connected in parallel, then what will be the equivalent resistance?
ii. A wire of resistance $12 \Omega$ is cut into three equal pieces and then twisted their ends together, then what will be the equivalent resistance?
iii. Three resistances are connected as shown. Calculate the equivalent resistance between $A$ and $B$
OR
Find the current in each resistance.
Answer
View full question & answer→i. The equivalent resistance in the parallel combination is lesser than the least value of the individual resistance.The equivalent resistance of parallel combinations
$\frac{1}{R p}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$
$\Rightarrow \mathrm{Rp}=\frac{8}{7} \Omega$
Thus equivalent resistance is less than $2 \Omega$.
ii. Resistance of each piece $=\frac{12}{3}=4 \Omega$
$\frac{1}{R_{p}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \Rightarrow R_{p}=\frac{4}{3} \Omega$
iii. All the three resistors are in paralle.
$\therefore \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{1}=\frac{1+2+6}{6}=\frac{9}{6} R_{P}=\frac{6}{9}=\frac{2}{3} \Omega$
OR
All are in parallel.
$\frac{1}{R_{p}}=\frac{1}{12} \times 4=\frac{1}{3} \Rightarrow R_{p}=3 \Omega$
$I=\frac{3}{3}=1 \mathrm{~A}$
So, current in each resistor $I^{\prime}=\frac{3}{12}=\frac{1}{4} \mathrm{~A}$
$\frac{1}{R p}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$
$\Rightarrow \mathrm{Rp}=\frac{8}{7} \Omega$
Thus equivalent resistance is less than $2 \Omega$.
ii. Resistance of each piece $=\frac{12}{3}=4 \Omega$
$\frac{1}{R_{p}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \Rightarrow R_{p}=\frac{4}{3} \Omega$
iii. All the three resistors are in paralle.
$\therefore \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{1}=\frac{1+2+6}{6}=\frac{9}{6} R_{P}=\frac{6}{9}=\frac{2}{3} \Omega$
OR
All are in parallel.
$\frac{1}{R_{p}}=\frac{1}{12} \times 4=\frac{1}{3} \Rightarrow R_{p}=3 \Omega$
$I=\frac{3}{3}=1 \mathrm{~A}$
So, current in each resistor $I^{\prime}=\frac{3}{12}=\frac{1}{4} \mathrm{~A}$