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CBSE Class 10 Physics · 11 questions · 26 marks
Electricity turns invisible charge flow into light, heat, and motion, and this chapter gives you the exact rules — Ohm's law, series and parallel combinations, and the heating effect — to calculate all of it. These are some of the most numerical-heavy, most scoring topics in the CBSE Class 10 Physics paper. Get comfortable with the formulas here and circuit problems stop being scary.
What is the SI unit of electrical resistance, and how is it related to V and I?
Answer
Ohm; R = V/I — resistance is the ratio of the potential difference across a conductor to the current flowing through it, as given by Ohm's law; its SI unit is the ohm (symbol Ω), equal to one volt per ampere.
Three resistors of 2 ohm, 3 ohm, and 6 ohm are connected in parallel. What is their equivalent resistance?
Answer
1 ohm — for resistors in parallel, 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R = 1 ohm. As expected, this equivalent resistance is smaller than the smallest individual resistor (2 ohm).
Why is a voltmeter designed to have very high resistance while an ammeter is designed to have very low resistance?
Answer
Because a voltmeter (connected in parallel) must draw negligible current to avoid disturbing the circuit, while an ammeter (connected in series) must not add noticeable resistance to the circuit — this ensures both instruments measure the circuit's true values without altering them.
A uniform copper wire is stretched so that its length becomes double, while its volume stays the same. How does its new resistance compare with the original resistance?
Answer
It becomes four times the original resistance — since volume (A x L) stays constant, doubling the length L halves the cross-sectional area A. As R = rho*L/A, doubling L and halving A both increase resistance, giving a combined factor of 2 x 2 = 4.
Assertion (A): When resistors are connected in series, the equivalent resistance of the combination is greater than the value of any individual resistor. Reason (R): In a series combination, the same current flows through every resistor, and the total potential difference across the combination equals the sum of the potential differences across each resistor.
Answer
Both A and R are true and R is the correct explanation of A — since the same current I flows through each resistor, V = IR1 + IR2 + IR3 = I(R1+R2+R3), so the equivalent resistance Rs = R1+R2+R3, which is necessarily greater than any single resistor in the sum.
Define electric current and state its SI unit.
Answer
Electric current is the rate of flow of electric charge through a conductor: I = Q/t, where Q is the charge (in coulombs) that flows past a given cross-section in time t (in seconds). Its SI unit is the ampere (A); a current of 1 A means 1 coulomb of charge flows per second.
A current of 0.5 A flows through a resistor of 8 ohm for 5 minutes. Calculate (i) the potential difference across the resistor, and (ii) the heat energy generated in it.
Answer
(i) V = IR = 0.5 A x 8 ohm = 4 V. (ii) Time t = 5 minutes = 300 s. Heat generated, H = I^2*R*t = (0.5)^2 x 8 x 300 = 0.25 x 8 x 300 = 600 J.
Why does the coiled heating element of an electric heater glow while the copper connecting wires carrying the same current do not?
Answer
The heating element (usually made of an alloy such as nichrome) has a much higher resistance and a higher melting point than the copper connecting wires. Since heat produced is given by H = I^2*R*t, for the same current and time far more heat is generated in the high-resistance element, raising it to a glowing temperature, while the low-resistance copper wires warm only slightly and do not glow.
Derive the expression for the equivalent resistance of three resistors R1, R2, and R3 connected in series, and state two characteristics of a series circuit. Then find the equivalent resistance and the current drawn from a 12 V battery when resistors of 4 ohm, 6 ohm, and 10 ohm are connected in series with it.
Answer
In a series combination, resistors R1, R2, R3 are joined end-to-end so the same current I flows through all of them. If V1, V2, V3 are the potential differences across R1, R2, R3, the total voltage supplied by the source is V = V1 + V2 + V3. By Ohm's law, V1 = IR1, V2 = IR2, V3 = IR3, so V = I(R1+R2+R3). If Rs is the single equivalent resistance drawing the same current I for the same voltage V, then V = I*Rs. Comparing the two expressions gives Rs = R1 + R2 + R3. Characteristics of a series circuit: (i) the same current flows through every component in the circuit; (ii) if any one component breaks or is disconnected, the entire circuit is broken and current stops flowing everywhere. Numerical: Rs = 4 + 6 + 10 = 20 ohm. Current drawn from the battery, I = V/Rs = 12/20 = 0.6 A.
State Joule's law of heating and derive the expression H = I^2*R*t using the definitions of current, potential difference, and work done. Give two practical applications of the heating effect of electric current.
Answer
Joule's law of heating states that the heat produced in a resistor is directly proportional to (i) the square of the current through it, (ii) its resistance, and (iii) the time for which the current flows: H = I^2*R*t. Derivation: let a current I flow through a resistor of resistance R for time t, under a potential difference V. In time t, the charge that flows is Q = It. The work done in moving this charge through the potential difference V is W = VQ = VIt. By Ohm's law, V = IR, so substituting gives W = (IR) x I x t = I^2*R*t. In a purely resistive conductor, this electrical work is converted entirely into heat, so the heat produced is H = W = I^2*R*t. Applications: (i) Electric heating appliances such as room heaters, electric irons, and toasters deliberately use this effect, employing high-resistance nichrome heating elements. (ii) Electric fuses use the heating effect as a safety device — a thin wire made of an alloy with a low melting point melts and disconnects the circuit when the current exceeds a safe limit, protecting the wiring and appliances from damage.
Read the following and answer the questions that follow: A household's monthly electricity bill lists two appliances: an electric iron rated "1000 W, 220 V" used for 30 minutes every day, and five LED bulbs, each rated "10 W, 220 V", used for 6 hours every day. The electricity board charges the household based on the total energy consumed in kilowatt-hour (commonly called a "unit") over the month. (a) Calculate the energy consumed by the electric iron in one day, in kWh. (b) Calculate the energy consumed by all five LED bulbs together in one day, in kWh. (c) What is meant by 1 unit (1 kWh) of electrical energy? (d) Suggest one reason why LED bulbs are more economical than old-style incandescent bulbs of similar brightness for long-term use.
Answer
(a) Energy = Power x time = 1000 W x 0.5 h = 500 Wh = 0.5 kWh. (b) Total power of the 5 bulbs = 5 x 10 W = 50 W. Energy = 50 W x 6 h = 300 Wh = 0.3 kWh. (c) 1 unit (1 kWh) of electrical energy is the energy consumed by an appliance of power 1000 W (1 kW) operated continuously for 1 hour. (d) LED bulbs consume far less power than incandescent bulbs for the same brightness, since they waste much less energy as heat; so, although they may cost more to buy initially, they consume less electrical energy over time, leading to lower electricity bills and reduced energy wastage.
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