To find the equivalent resistance \( R_{\text{eq}} \) of resistors connected in parallel, we use the formula:
\[\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]
where \( R_1 = 2\, \Omega \), \( R_2 = 3\, \Omega \), and \( R_3 = 6\, \Omega \).
Substituting the given values into the formula:
\[\frac{1}{R_{\text{eq}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}\]
Finding a common denominator (6) for the fractions:
\[\frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6}, \quad \frac{1}{6} = \frac{1}{6}\]
Adding the fractions:
\[\frac{1}{R_{\text{eq}}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1\]
Thus:
\[R_{\text{eq}} = \frac{1}{1} = 1\, \Omega\]
Therefore, the equivalent resistance of the combination is \( 1\, \Omega \).
The current passing through the battery in the given circuit, is:
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :