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 calculating the heat dissipated by the 4-ohm resistor in a series circuit with a 6 V battery, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Resistors and Battery**: We have two resistors: R1 = 2 Ω and R2 = 4 Ω connected in series to a 6 V battery.

2. **Calculate the Total Resistance**: In a series circuit, the total resistance (R_total) is the sum of the individual resistances.
\[ R_{\text{total}} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega \]

3. **Use Ohm's Law to Find Current**: According to Ohm's Law, the current (I) flowing through the circuit can be calculated using the formula:
\[ V = I \times R_{\text{total}} \implies I = \frac{V}{R_{\text{total}}} \]
Substituting the values:
\[ I = \frac{6 \, V}{6 \, \Omega} = 1 \, A \]

4. **Calculate Heat Dissipated by the 4 Ω Resistor**: The heat (H) dissipated by a resistor can be calculated using the formula:
\[ H = I^2 \times R \times T \]
Here, we will use the resistance of the 4 Ω resistor (R = 4 Ω) and the time (T = 5 s):
\[ H = (1 \, A)^2 \times 4 \, \Omega \times 5 \, s = 1 \times 4 \times 5 = 20 \, J \]

5. **Final Answer**: The heat dissipated by the 4 Ω resistor in 5 seconds is:
\[ H = 20 \, J \]