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} $.
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)