Let's simplify the given circuit.
The circuit consists of three resistors, each with a resistance of $ r/3 $. Let's label them $ R_1 $, $ R_2 $, and $ R_3 $.
We know:
$ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $
$ \Rightarrow \frac{1}{R_p} = \frac{3}{r} + \frac{3}{r} + \frac{3}{r} = \frac{9}{r} $
$ \Rightarrow R_p = \frac{r}{9} $
Final Answer:
The final answer is $ \boxed{r/9} $.
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 :
The current passing through the battery in the given circuit, is:
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: