Question

The ratio of heat dissipated per second through the resistance 5 ohm and 10 ohm in the circuit given below is :

• A. 1:2
• B. 2:1
• C. 4:1
• D. 1:1

Answer 1

The Correct Option is B

Given circuit:
The \( 5 \, \Omega \) and \( 10 \, \Omega \) resistors are connected in parallel.

Step 1: Calculating the Equivalent Resistance
The equivalent resistance \( R_{\text{eq}} \) of the parallel combination is given by:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{5} + \frac{1}{10}. \]

Calculating:

\[ \frac{1}{R_{\text{eq}}} = \frac{2}{10} + \frac{1}{10} = \frac{3}{10} \implies R_{\text{eq}} = \frac{10}{3} \, \Omega. \]

Step 2: Current Division in Parallel Resistors
Let \( i_1 \) be the current through the \( 5 \, \Omega \) resistor and \( i_2 \) be the current through the \( 10 \, \Omega \) resistor. By the current division rule:

\[ \frac{i_1}{i_2} = \frac{R_2}{R_1} = \frac{10}{5} = 2. \]

Thus, \( i_1 = 2i_2 \).

Step 3: Calculating the Power Dissipated
The power dissipated \( P \) in a resistor is given by:

\[ P = i^2R. \]

The ratio of the power dissipated in the \( 5 \, \Omega \) resistor to the \( 10 \, \Omega \) resistor is:

\[ \frac{P_1}{P_2} = \frac{i_1^2R_1}{i_2^2R_2} = \left( \frac{i_1}{i_2} \right)^2 \times \frac{R_1}{R_2}. \]

Substituting the values:

\[ \frac{P_1}{P_2} = (2)^2 \times \frac{5}{10} = 4 \times \frac{1}{2} = 2. \]

Therefore, the ratio of heat dissipated per second through the \( 5 \, \Omega \) and \( 10 \, \Omega \) resistors is \( 2:1 \).