Question

Two resistors R and 2R are connected in series in an electric circuit. The thermal energy developed in R and 2R will be in ratio

• A. 1:2
• B. 1:4
• C. 2:1
• D. 4:1

Hint:

For resistors in **series**, current is constant, so use \(H = I^2Rt\), which means \(H \propto R\). For resistors in **parallel**, voltage is constant, so use \(H = \frac{V^2}{R}t\), which means \(H \propto \frac{1}{R}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When current flows through a resistor, electrical energy is converted into thermal energy. This is known as Joule heating.
Step 2: Key Formula or Approach:
The thermal energy (\(H\)) developed in a resistor is given by Joule's law of heating:
\[ H = I^2 R t \] where \(I\) is the current, \(R\) is the resistance, and \(t\) is the time for which the current flows.
Step 3: Detailed Explanation:
The two resistors, \(R_1 = R\) and \(R_2 = 2R\), are connected in series.
In a series circuit, the current flowing through each component is the same. Let this current be \(I\).
The time \(t\) for which the current flows is also the same for both resistors.
Therefore, the thermal energy developed in each resistor is directly proportional to its resistance (\(H \propto R\)).
Let \(H_1\) be the energy developed in resistor \(R\) and \(H_2\) be the energy developed in resistor \(2R\).
\[ H_1 = I^2 R t \] \[ H_2 = I^2 (2R) t \] To find the ratio of the thermal energy, we divide \(H_1\) by \(H_2\):
\[ \frac{H_1}{H_2} = \frac{I^2 R t}{I^2 (2R) t} \] The terms \(I^2\) and \(t\) cancel out, leaving:
\[ \frac{H_1}{H_2} = \frac{R}{2R} = \frac{1}{2} \] So, the ratio of the thermal energy is \(1:2\).
Step 4: Final Answer:
The ratio of thermal energy developed in R and 2R is 1:2. Therefore, option (A) is correct.