`6Omega` and `12Omega` resistors are connected in parallel. This combination is connected in series with 10 V battery and `6 Omega` resistor. What is the potential differnce between the terminals of the `12Omega` resistor ?

Options

A. 4 V
B. 16 V
C. 2
D. 8 V

Answer

To solve the problem step by step, let's break it down:

### Step 1: Identify the Circuit Configuration
We have a `6Ω` resistor and a `12Ω` resistor connected in parallel. This combination is then connected in series with a `6Ω` resistor and a `10V` battery.

### Step 2: Calculate the Equivalent Resistance of the Parallel Combination
The formula for the equivalent resistance \( R_{eq} \) of two resistors \( R_1 \) and \( R_2 \) in parallel is given by:

\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Substituting \( R_1 = 6Ω \) and \( R_2 = 12Ω \):

\[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{12} \]

Finding a common denominator (which is 12):

\[ \frac{1}{R_{eq}} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \]

Thus,

\[ R_{eq} = 4Ω \]

### Step 3: Total Resistance in the Circuit
Now, we have the equivalent resistance of the parallel combination \( R_{eq} = 4Ω \) in series with the internal resistance of the battery \( R_{internal} = 6Ω \). The total resistance \( R_{total} \) is:

\[ R_{total} = R_{eq} + R_{internal} = 4Ω + 6Ω = 10Ω \]

### Step 4: Calculate the Current in the Circuit
Using Ohm's law, the current \( I \) flowing through the circuit can be calculated as:

\[ I = \frac{V}{R_{total}} = \frac{10V}{10Ω} = 1A \]

### Step 5: Determine the Voltage Across the Parallel Resistors
Since the `6Ω` and `12Ω` resistors are in parallel, they share the same voltage. The voltage across the parallel combination \( V_{parallel} \) can be calculated using Ohm's law again:

\[ V_{parallel} = I \times R_{eq} = 1A \times 4Ω = 4V \]

### Step 6: Calculate the Current Through the `12Ω` Resistor
In a parallel circuit, the current divides inversely proportional to the resistance. The total current \( I \) is 1A, and we can find the current through the `12Ω` resistor \( I_{12} \):

Using the current division rule:

\[ I_{12} = I \times \frac{R_{other}}{R_{1} + R_{2}} = 1A \times \frac{6Ω}{6Ω + 12Ω} = 1A \times \frac{6}{18} = \frac{1}{3}A \]

### Step 7: Calculate the Voltage Across the `12Ω` Resistor
Now, we can find the potential difference \( V_{12} \) across the `12Ω` resistor using Ohm's law:

\[ V_{12} = I_{12} \times R_{12} = \frac{1}{3}A \times 12Ω = 4V \]

### Conclusion
The potential difference between the terminals of the `12Ω` resistor is **4 volts**.

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