Three equal resistances each of `3Omega` are in series and connected to a cell of internal resistance one ohm. If these resistance are in parallel and connected to the same cell, then the ratio of the respective currents through the electric circuits in the two cases is

Options

A. `1//8`
B. `1//7`
C. `1//5`
D. `1//3`

Explanation

To solve the problem, we need to calculate the currents in two different configurations of resistors connected to a cell with internal resistance. Let's break it down step by step.

### Step 1: Calculate the equivalent resistance when the resistors are in series.

When resistors are connected in series, the total or equivalent resistance (R_s) is simply the sum of the individual resistances.

Given:
- Each resistance, R = 3Ω
- Number of resistors = 3

The equivalent resistance in series is:
\[ R_s = R_1 + R_2 + R_3 = 3Ω + 3Ω + 3Ω = 9Ω \]

### Step 2: Include the internal resistance of the cell.

The internal resistance (r) of the cell is given as 1Ω. The total resistance in the circuit when the resistors are in series is:
\[ R_{total, series} = R_s + r = 9Ω + 1Ω = 10Ω \]

### Step 3: Calculate the current through the circuit when resistors are in series.

Using Ohm's law, the current (I_s) through the circuit can be calculated as:
\[ I_s = \frac{E}{R_{total, series}} \]
Where E is the electromotive force (emf) of the cell. Thus:
\[ I_s = \frac{E}{10Ω} \]

### Step 4: Calculate the equivalent resistance when the resistors are in parallel.

When resistors are connected in parallel, the equivalent resistance (R_p) can be calculated using the formula:
\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Substituting the values:
\[ \frac{1}{R_p} = \frac{1}{3Ω} + \frac{1}{3Ω} + \frac{1}{3Ω} = \frac{3}{3Ω} = 1 \]
Thus,
\[ R_p = 1Ω \]

### Step 5: Include the internal resistance of the cell for the parallel case.

The total resistance in the circuit when the resistors are in parallel is:
\[ R_{total, parallel} = R_p + r = 1Ω + 1Ω = 2Ω \]

### Step 6: Calculate the current through the circuit when resistors are in parallel.

Using Ohm's law again, the current (I_p) through the circuit can be calculated as:
\[ I_p = \frac{E}{R_{total, parallel}} \]
Thus:
\[ I_p = \frac{E}{2Ω} \]

### Step 7: Calculate the ratio of the currents.

Now, we can find the ratio of the currents in the two cases:
\[ \text{Ratio} = \frac{I_s}{I_p} = \frac{\frac{E}{10Ω}}{\frac{E}{2Ω}} \]
The E cancels out:
\[ \text{Ratio} = \frac{2}{10} = \frac{1}{5} \]

### Final Answer:
The ratio of the respective currents through the electric circuits in the two cases is \( \frac{1}{5} \).

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