Question

The current through the circuit shown in figure is $1A$. If each of $4\Omega$ the resistors is replaced by $2\Omega$ resistor, the current in circuit will become nearly

• A. 1.11 A
• B. 1.25 A
• C. 1.34 A
• D. 1.67 A

Answer 1

Answer: (A)

In the circuit shown total external resistance $R=2\Omega +$ parallel combination of two $4\Omega$ resistors + parallel combination of three $15\Omega$ resistors.
$=2+\frac{4}{2}+\frac{15}{3}=2+2+5=9\Omega$
As $E=10V$ and $i=1A$, hence internal resistance $r$ of the cell should have a value given by
$E=i(R+r)$
$r=\frac{E}{i}-R=\frac{10}{1}-9=1\Omega$
If $4\Omega$ resistors are replaced by $2\Omega$ resistors, then as before $R^{\prime }=2+\frac{2}{2}+\frac{15}{3}=2+1+5=8\Omega$
$\therefore$ New circuit current $i^{\prime }=\frac{E}{R^{\prime }+r}=\frac{10}{8+1}=1.11A$