Electricity — Karnataka (SSLC) Class 10 Science Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 10 Science solutions for "Electricity" — important questions with detailed answers, download PDF for board exam preparation.
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Electricity — Textbook Questions Answered (7)
Q1: State Ohm's law and define electric resistance. What is the unit of resistance?
Ohm's law: V = IR, where voltage (V) is directly proportional to current (I) and resistance (R). Resistance is the opposition offered by a conductor to the flow of electric current. Unit: Ohm (Ω). A conductor has 1 Ω resistance if 1 A current flows when 1 V is applied.
Q2: What is resistivity? How does it differ from resistance? Give the formula.
Resistivity (ρ) is the property of a material that determines its resistance; it's intrinsic to the material. Resistance (R) depends on resistivity, length, and area. Formula: R = ρL/A, where L = length, A = cross-sectional area. Unit of ρ: Ω⋅m. Materials with lower ρ are better conductors.
Q3: Calculate the current flowing through a 5 Ω resistor when 10 V is applied across it.
Step 1: Given V = 10 V, R = 5 Ω
Step 2: Using Ohm's law: V = IR
Step 3: I = V/R = 10/5 = 2 A
Step 4: Current flowing through the resistor is 2 amperes
Q4: What is electric power and energy? How are they related? Give formulas.
Power (P) is the rate of energy consumption: P = VI = I²R = V²/R (watts). Energy (E) is total work done over time: E = Pt = VIt (joules or kWh). 1 unit of electricity = 1 kWh = 3.6 × 10⁶ J. A 100 W bulb uses 0.1 kW; run 10 h = 1 kWh energy.
Q5: Explain series and parallel circuits. Compare their properties.
Series: components connected one after another. Current same everywhere, voltage divides, Rₜ = R₁+R₂+R₃. Parallel: components connected across same points. Voltage same, current divides, 1/Rₜ = 1/R₁+1/R₂+1/R₃. Series used for switches; parallel for home wiring (independent control).
Q6: A copper wire of length 2 m and cross-sectional area 1 mm² (resistivity 1.6 × 10⁻⁸ Ω⋅m) is connected across 5 V. Find current.
Step 1: L = 2 m, A = 1 mm² = 10⁻⁶ m², ρ = 1.6 × 10⁻⁸ Ω⋅m, V = 5 V
Step 2: R = ρL/A = (1.6 × 10⁻⁸ × 2)/(10⁻⁶) = 0.032 Ω
Step 3: I = V/R = 5/0.032 = 156.25 A
Step 4: Current is approximately 156 amperes
Q7: Three resistors 2 Ω, 3 Ω, 6 Ω are connected in parallel across 12 V. Find equivalent resistance and total current.
Step 1: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 6 Ω, V = 12 V
Step 2: 1/Rₑq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1
Step 3: Rₑq = 1 Ω
Step 4: Total current I = V/Rₑq = 12/1 = 12 A
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