Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Resistances $1 \, \Omega , \, 2 \, \Omega $ and $3 \, \Omega $ are connected to form a triangle. If a $1.5 \, V$ cell of negligible internal resistance is connected across the $3 \, \Omega $ resistor, the current flowing through this resistor will be

NTA AbhyasNTA Abhyas 2020Current Electricity

Solution:

Equivalent resistance between A and B = series combination of $1 \Omega $ and $2 \Omega $ in parallel with $3 \Omega $ resistor.
Solution
$\frac{1}{R} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$ or $R = 1.5 \Omega $ .
$\therefore $ Current in the circuits , $I=\frac{V}{R}=\frac{1 . 5 }{1 . 5}=1A$
Since the resistance in arm ACB = resistance in arm $A B = 3 \Omega $ , the current divided equally in the two arms. Hence the current through the $3 \Omega $ resistor = $\frac{I}{2}=0.5A$ .