Science (086) — CBSE Class 10 board question

Since the exact circuit diagram is not visible, I will solve using the standard textbook combination (Example 11.9): R₁ = 10 Ω, R₂ = 40 Ω in parallel, and R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω in parallel, connected in series.

Step 1: Parallel combination of R₁ and R₂:
$$\frac{1}{R'} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40} \Rightarrow R' = 8\ \Omega$$

Step 2: Parallel combination of R₃, R₄, R₅:
$$\frac{1}{R''} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{6}{60} \Rightarrow R'' = 10\ \Omega$$

Step 3: Total resistance (series):
$$R = R' + R'' = 8 + 10 = \boxed{18\ \Omega}$$

Source: Chapter 11, Section 11.6 (Resistance of a System of Resistors), Example 11.9

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• Examiners want clear stepwise working — identify series/parallel groups first, then combine.
• The standard CBSE question on mixed circuits uses this exact example (R' = 8 Ω, R'' = 10 Ω, total = 18 Ω).
• Always state the formula used ($\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+…$ for parallel; $R_s = R_1+R_2+…$ for series) before substituting.
• If your circuit has different values, apply the same method: simplify parallel groups first, then add in series.