A cell of e.m.f. 2V and internal resistance `0.5Omega` is connected across a resistor R. The current that flows is same as that, when a cell of e.m.f. 1.5V and internal resistance `0.3Omega` is connected across the same resistor. Then

Options

A. `R=0.3Omega`
B. `R=0.6Omega`
C. `R=0.5 Omega`
D. `R=0.75Omega`

Explanation

To solve the problem step by step, we will use Ohm's law and the concept of equivalent resistance in series circuits.

### Step 1: Identify the components of the first circuit
We have a cell with an e.m.f. \( E_1 = 2V \) and internal resistance \( r_1 = 0.5 \Omega \) connected across a resistor \( R \). The current \( I \) flowing through the circuit can be expressed using Ohm's law:

\[ I = \frac{E_1}{R + r_1} \]

Substituting the values:

\[ I = \frac{2}{R + 0.5} \]

### Step 2: Identify the components of the second circuit
In the second circuit, we have a cell with an e.m.f. \( E_2 = 1.5V \) and internal resistance \( r_2 = 0.3 \Omega \) connected across the same resistor \( R \). The current \( I \) flowing through this circuit is:

\[ I = \frac{E_2}{R + r_2} \]

Substituting the values:

\[ I = \frac{1.5}{R + 0.3} \]

### Step 3: Set the currents equal
Since the current \( I \) is the same in both circuits, we can set the two equations equal to each other:

\[ \frac{2}{R + 0.5} = \frac{1.5}{R + 0.3} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:

\[ 2(R + 0.3) = 1.5(R + 0.5) \]

### Step 5: Expand both sides
Expanding both sides results in:

\[ 2R + 0.6 = 1.5R + 0.75 \]

### Step 6: Rearrange the equation
Rearranging the equation to isolate \( R \):

\[ 2R - 1.5R = 0.75 - 0.6 \]

This simplifies to:

\[ 0.5R = 0.15 \]

### Step 7: Solve for \( R \)
Now, divide both sides by 0.5 to find \( R \):

\[ R = \frac{0.15}{0.5} = 0.3 \Omega \]

### Conclusion
The value of the resistor \( R \) is \( 0.3 \Omega \).

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