Question

Three conductors draw currents of 1A, 2A and 3A respectively, when connected in a turn across a battery. If they are connected in series and the combination is connected across the same battery, the current drawn will be:
A. ${\displaystyle \frac{6}{11}}A$
B. ${\displaystyle \frac{3}{7}}A$
C. ${\displaystyle \frac{4}{7}}A$
D. ${\displaystyle \frac{5}{7}}A$

Answer 1

Hint: In electronics and electromagnetism, the electrical resistance of an object is a measure of its opposition to the flow of electric current. The resistance of an object depends in large part on the material it is made of. Objects made of electrical insulators like rubber tend to have very high resistance and low conductivity, while objects made of electrical conductors like metals tend to have very low resistance and high conductivity.
One of the most important law in electronics is Ohm’s law. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. Introducing the constant of proportionality, the resistance.
$I\alpha V\Rightarrow I=RV$

Complete step by step solution:
Let the potential of the battery be V volts.
A/q to Ohm’s law $R={\displaystyle \frac{V}{I}}$
$\because$Conductor 1 draws 1A
$\therefore$Resistance of conductor 1 $R1={\displaystyle \frac{V}{1}}$
$\because$ Conductor 2 draws 2A
$\therefore$ Resistance of conductor 2 $R2={\displaystyle \frac{V}{2}}$


$\because$ Conductor 3 draws 3A
$\therefore$ Resistance of conductor 3$R3={\displaystyle \frac{V}{3}}$
Now, they are connected in series therefore effective resistance will be simply the sum of all three resistances.
$\begin{array}{c}\mathrm{Re}q=R1+R2+R3\\ \Rightarrow \mathrm{Re}q={\displaystyle \frac{V}{1}}+{\displaystyle \frac{V}{2}}+{\displaystyle \frac{V}{3}}\\ \Rightarrow \mathrm{Re}q={\displaystyle \frac{6V+3V+2V}{6}}\\ \Rightarrow \mathrm{Re}q={\displaystyle \frac{11V}{6}}\\ \end{array}$
  When all are connected across the battery the current drawn will be
: $\begin{array}{c}I={\displaystyle \frac{V}{\mathrm{Re}q}}\\ \Rightarrow I={\displaystyle \frac{V}{{\displaystyle \frac{11V}{6}}}}\\ \Rightarrow I={\displaystyle \frac{6}{11}}\end{array}$


Thus the required current drawn will be = ${\displaystyle \frac{6}{11}}A$


Note:For solving these types of questions first of all identify the elements that you need to find and then proceed step by step. Also, be careful while solving and try to avoid the confusion between the resistances in series and parallel.