The ratio of heat dissipated per second through the resistance 5 ohm and 10 ohm in the circuit given below is :
- 1:2
- 2:1
- 4:1
- 1:1
The Correct Option is B
Approach Solution - 1
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 \( 2:1 \).
Approach Solution -2
E = 10 V
R₁ = 20 Ω (in series)
R₂ = 5 Ω (in parallel with 10 Ω)
R₃ = 10 Ω
Step 2: Simplify the parallel part (5 Ω and 10 Ω).
The equivalent resistance of parallel resistors is given by:
\[ \frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} = \frac{2 + 1}{10} = \frac{3}{10} \] \[ R_p = \frac{10}{3} \, \Omega \]
Step 3: Find the total resistance in the circuit.
Total resistance, R = 20 + Rₚ = 20 + 10/3 = 70/3 ≈ 23.33 Ω
Step 4: Find total current from the battery.
\[ I = \frac{V}{R} = \frac{10}{70/3} = \frac{30}{70} = \frac{3}{7} \, \text{A} \]
Step 5: Find voltage across the parallel combination.
Voltage across parallel branch (Vₚ) = I × Rₚ = (3/7) × (10/3) = 10/7 V
Step 6: Find current through each resistor in parallel.
For 5 Ω resistor: \( I₁ = \frac{V_p}{5} = \frac{10/7}{5} = \frac{2}{7} \, \text{A} \)
For 10 Ω resistor: \( I₂ = \frac{V_p}{10} = \frac{10/7}{10} = \frac{1}{7} \, \text{A} \)
Step 7: Find heat dissipated per second (power) in each resistor.
Power dissipated, \( P = I^2 R \)
For 5 Ω resistor: \( P₁ = (\frac{2}{7})^2 \times 5 = \frac{4}{49} \times 5 = \frac{20}{49} \)
For 10 Ω resistor: \( P₂ = (\frac{1}{7})^2 \times 10 = \frac{1}{49} \times 10 = \frac{10}{49} \)
Step 8: Ratio of heat dissipated per second.
\[ P₁ : P₂ = \frac{20}{49} : \frac{10}{49} = 2 : 1 \]
Final Answer: 2 : 1
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