Question

A circuit is made using $R_1,R_2,R_3,R_4$ and a battery as shown in the following figure. Find the equivalent resistance of the given circuit and the current passing through $R_3$

• A. $3\Omega ,\frac{1}{3}A$
• B. $\frac{1}{3}\Omega ,27A$
• C. $\frac{2}{3}\Omega ,\frac{21}{2}A$
• D. $\frac{1}{3}\Omega ,\frac{21}{2}A$
• E. $\frac{2}{3}\Omega ,\frac{21}{3}A$

Answer 1

Answer: (B)

The circuit can be redrawn as shown in the figure below,

So, the equivalent resistance of the three parallel branches is given as,
$R_{eq}=\frac{1}{\frac{1}{0.5+0.5}+\frac{1}{1}+\frac{1}{1}}=\frac{1}{3}\Omega$
Now, from Ohm's law
$I=\frac{V}{R}=\frac{27}{1/3}=27\times 3=18A$
Here in the figure, 3 branches are of equal resistance, so current will divide equally in each branch, So, current in $R_3$,
$I_{R_3}=\frac{27\times 3}{3}=\frac{81}{3}=27A$