Combination Of Resistors Series And Parallel

To find the value of $n$ , the factor by which current increases when the resistors are connected in parallel compared to series, let's analyze both configurations. When resistors are connected in series, the total resistance $R_s$ is:

$R_s%20%3D%2010R$

The current $I_s$ through the series circuit connected to a battery of emf $E$ is given by Ohm's Law:

$I_s%20%3D%20%5Cfrac%7BE%7D%7BR_s%7D%20%3D%20%5Cfrac%7BE%7D%7B10R%7D$

For resistors connected in parallel, the total resistance $R_p$ is:

$%5Cfrac%7B1%7D%7BR_p%7D%20%3D%20%5Cfrac%7B1%7D%7BR%7D%20%2B%20%5Cfrac%7B1%7D%7BR%7D%20%2B%20%5Ccdots%20%2B%20%5Cfrac%7B1%7D%7BR%7D%20%3D%20%5Cfrac%7B10%7D%7BR%7D$

Simplifying, we get:

$R_p%20%3D%20%5Cfrac%7BR%7D%7B10%7D$

The current $I_p$ through the parallel circuit is:

$I_p%20%3D%20%5Cfrac%7BE%7D%7BR_p%7D%20%3D%20%5Cfrac%7BE%7D%7B%5Cfrac%7BR%7D%7B10%7D%7D%20%3D%20%5Cfrac%7B10E%7D%7BR%7D$

The ratio of currents $n$ is:

$n%20%3D%20%5Cfrac%7BI_p%7D%7BI_s%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B10E%7D%7BR%7D%7D%7B%5Cfrac%7BE%7D%7B10R%7D%7D%20%3D%20100$

Therefore, the value of $n$ is 100.

Step 1: Understanding the Concept:
This is a circuit analysis problem. First, we need to determine the resistance of each piece of wire. Then, we analyze the circuit formed by the triangle of resistors to find the equivalent external resistance. Finally, we use Ohm's law for the entire circuit, including the internal resistance of the cell, to find the total current.

Step 2: Key Formula or Approach:
1. For a uniform wire, resistance is proportional to length ( $R%20%5Cpropto%20L$ ). 2. Resistors in series: $R_%7Bseries%7D%20%3D%20R_1%20%2B%20R_2%20%2B%20%5Cdots$ 3. Resistors in parallel: $%5Cfrac%7B1%7D%7BR_%7Bparallel%7D%7D%20%3D%20%5Cfrac%7B1%7D%7BR_1%7D%20%2B%20%5Cfrac%7B1%7D%7BR_2%7D%20%2B%20%5Cdots$ 4. Ohm's law for a complete circuit: $I%20%3D%20%5Cfrac%7B%5Cmathcal%7BE%7D%7D%7BR_%7Bext%7D%20%2B%20r%7D$ , where $R_%7Bext%7D$ is the equivalent external resistance and $r$ is the internal resistance.

Step 3: Detailed Explanation:
Part 1: Find the resistances of the three pieces.
Total resistance $R_%7Btotal%7D%20%3D%2012%20%5C%2C%20%5COmega$ . The lengths are in the ratio 1:2:3. Since $R%20%5Cpropto%20L$ , the resistances will also be in the ratio 1:2:3. Let the resistances be $R_1%2C%20R_2%2C%20R_3$ . $R_1%20%3D%20%5Cleft(%5Cfrac%7B1%7D%7B1%2B2%2B3%7D%5Cright)%20%5Ctimes%2012%20%5C%2C%20%5COmega%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%5Ctimes%2012%20%3D%202%20%5C%2C%20%5COmega$ $R_2%20%3D%20%5Cleft(%5Cfrac%7B2%7D%7B1%2B2%2B3%7D%5Cright)%20%5Ctimes%2012%20%5C%2C%20%5COmega%20%3D%20%5Cfrac%7B2%7D%7B6%7D%20%5Ctimes%2012%20%3D%204%20%5C%2C%20%5COmega$ $R_3%20%3D%20%5Cleft(%5Cfrac%7B3%7D%7B1%2B2%2B3%7D%5Cright)%20%5Ctimes%2012%20%5C%2C%20%5COmega%20%3D%20%5Cfrac%7B3%7D%7B6%7D%20%5Ctimes%2012%20%3D%206%20%5C%2C%20%5COmega$ The highest resistance is $R_3%20%3D%206%20%5C%2C%20%5COmega$ . Part 2: Find the equivalent external resistance ( $R_%7Bext%7D$ ).
The three resistors form a triangle. The cell is connected across the highest resistor, $R_3$ . This means that resistors $R_1$ and $R_2$ are in series with each other, and this series combination is in parallel with $R_3$ . Resistance of the series combination: $R_%7B12%7D%20%3D%20R_1%20%2B%20R_2%20%3D%202%20%5C%2C%20%5COmega%20%2B%204%20%5C%2C%20%5COmega%20%3D%206%20%5C%2C%20%5COmega$ . Now, $R_%7B12%7D$ is in parallel with $R_3$ . The equivalent external resistance $R_%7Bext%7D$ is: $%5Cfrac%7B1%7D%7BR_%7Bext%7D%7D%20%3D%20%5Cfrac%7B1%7D%7BR_%7B12%7D%7D%20%2B%20%5Cfrac%7B1%7D%7BR_3%7D%20%3D%20%5Cfrac%7B1%7D%7B6%20%5C%2C%20%5COmega%7D%20%2B%20%5Cfrac%7B1%7D%7B6%20%5C%2C%20%5COmega%7D%20%3D%20%5Cfrac%7B2%7D%7B6%20%5C%2C%20%5COmega%7D%20%3D%20%5Cfrac%7B1%7D%7B3%20%5C%2C%20%5COmega%7D$ $R_%7Bext%7D%20%3D%203%20%5C%2C%20%5COmega$ Part 3: Calculate the total current.
Given data for the cell:
EMF, $%5Cmathcal%7BE%7D%20%3D%208%20%5C%2C%20%5Ctext%7BV%7D$ .
Internal resistance, $r%20%3D%205%20%5C%2C%20%5COmega$ .
Using Ohm's law for the complete circuit: $I%20%3D%20%5Cfrac%7B%5Cmathcal%7BE%7D%7D%7BR_%7Bext%7D%20%2B%20r%7D%20%3D%20%5Cfrac%7B8%20%5C%2C%20%5Ctext%7BV%7D%7D%7B3%20%5C%2C%20%5COmega%20%2B%205%20%5C%2C%20%5COmega%7D%20%3D%20%5Cfrac%7B8%20%5C%2C%20%5Ctext%7BV%7D%7D%7B8%20%5C%2C%20%5COmega%7D%20%3D%201%20%5C%2C%20%5Ctext%7BA%7D$

Step 4: Final Answer:
The current through the circuit (drawn from the cell) is 1 A.

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Combination Of Resistors Series And Parallel

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