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

Answer 1

Answer: (B)

$\frac{1}{3} \Omega , 27 A$

Answer 2

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

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