Question:
Three resistances of values 2Ω, 3Ω, and 6Ω are to be connected to produce an effective resistance of 4Ω. This can be done by connecting:
Three resistances of values 2Ω, 3Ω, and 6Ω are to be connected to produce an effective resistance of 4Ω. This can be done by connecting:
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For resistances in series, the total resistance is the sum of individual resistances. For resistances in parallel, use the formula \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \).
Updated On: Apr 28, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Parallel combination of resistors
The 3 Ω and 6 Ω resistors are in parallel. Their equivalent resistance is:
\[ R_p = \frac{3 \times 6}{3 + 6} \] \[ R_p = \frac{18}{9} = 2\ \Omega \]
Step 2: Series combination
This equivalent resistance is in series with a 2 Ω resistor:
\[ R_{\text{total}} = 2 + 2 \] \[ R_{\text{total}} = 4\ \Omega \]
Final Answer:
\( R_{\text{total}} = 4\ \Omega \)
The 3 Ω and 6 Ω resistors are in parallel. Their equivalent resistance is:
\[ R_p = \frac{3 \times 6}{3 + 6} \] \[ R_p = \frac{18}{9} = 2\ \Omega \]
Step 2: Series combination
This equivalent resistance is in series with a 2 Ω resistor:
\[ R_{\text{total}} = 2 + 2 \] \[ R_{\text{total}} = 4\ \Omega \]
Final Answer:
\( R_{\text{total}} = 4\ \Omega \)
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