In an electrical circuit, two resistors of `2 Omega and 4 Omega` respectively are connected in series to a `6 V` battery. The heat dissipated by the `4 Omega` resistor in `5 s` will be :

Options

A. 5 J
B. 10 J
C. 20 J
D. 30 J

Explanation

To solve the problem of finding the heat dissipated by the 4-ohm resistor in a series circuit with a 6V battery, we can follow these steps:

### Step 1: Identify the given values
- Resistor 1, \( R_1 = 2 \, \Omega \)
- Resistor 2, \( R_2 = 4 \, \Omega \)
- Voltage of the battery, \( V = 6 \, V \)
- Time, \( t = 5 \, s \)

### Step 2: Calculate the equivalent resistance
Since the resistors are connected in series, the total or equivalent resistance \( R_{eq} \) can be calculated as:
\[ R_{eq} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega \]

### Step 3: Calculate the current in the circuit
Using Ohm's Law, the current \( I \) can be calculated using the formula:
\[ I = \frac{V}{R_{eq}} = \frac{6 \, V}{6 \, \Omega} = 1 \, A \]

### Step 4: Calculate the heat dissipated by the 4-ohm resistor
The heat \( H \) dissipated by a resistor can be calculated using the formula:
\[ H = I^2 \cdot R \cdot t \]
For the 4-ohm resistor:
\[ H = (1 \, A)^2 \cdot (4 \, \Omega) \cdot (5 \, s) = 1 \cdot 4 \cdot 5 = 20 \, J \]

### Final Answer
The heat dissipated by the 4-ohm resistor in 5 seconds is \( 20 \, J \).

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