CBSE Class 10 Physics Electricity MCQs Set 15

Answer: (a) directly proportional to the potential difference across it

Explanation:
1. Ohm's law: V \( \propto \) I, or V/I = R, a constant for a given conductor at constant temperature.
2. So the V-I graph is a straight line through the origin.

Teacher's Note:
1. R is called the resistance of the conductor.
2. Conductors that obey Ohm's law are called ohmic.

Answer: (b) in parallel with the resistance coil

Explanation:
1. The voltmeter measures the potential difference between two points, so it is joined across (in parallel with) the coil.
2. The ammeter, which measures the current through the coil, is connected in series.

Teacher's Note:
1. A voltmeter has a very high resistance.
2. An ammeter has a very low resistance.

Answer: (c) its resistance is very small and it must carry the whole current without changing it

Explanation:
1. A good ammeter has almost zero resistance, so inserting it in series does not change the current to be measured.
2. Connected in parallel it would offer a short path and take a large current that could damage it.

Teacher's Note:
1. The voltmeter has a high resistance for the opposite reason.
2. Check the zero error of the meters before use.

Answer: (d) 5 \( \Omega \)

Explanation:
1. R = V/I = 2.0/0.4.
2. R = 5 \( \Omega \).

Teacher's Note:
1. Resistance is measured in ohms.
2. R is the slope of the V-I graph.

Answer: (c) 5 \( \Omega \)

Explanation:
1. The slope = \( \Delta \)V/\( \Delta \)I = (3.0 \( - \) 1.0)/(0.6 \( - \) 0.2) = 2.0/0.4 = 5.
2. The slope of the V-I graph is the resistance, 5 \( \Omega \).

Teacher's Note:
1. The graph is a straight line through the origin.
2. V/I is the same in all three readings.

Answer: (d) the heating of the coil changes its resistance

Explanation:
1. Ohm's law holds only at constant temperature.
2. A large current heats the wire, and its resistance rises, so V/I would no longer be constant.

Teacher's Note:
1. Use the key only while taking a reading.
2. A rheostat controls the current.

Answer: (d) vary the current and potential difference

Explanation:
1. A rheostat is a variable resistance, so moving its slider changes the current.
2. This gives a set of different readings of I and V.

Teacher's Note:
1. A set of readings gives a more reliable value of R.
2. The rheostat is connected in series.

Answer: (c) 10 \( \Omega \)

Explanation:
1. In series, R = r\( _1 \) + r\( _2 \) = 4 + 6.
2. R = 10 \( \Omega \).

Teacher's Note:
1. The equivalent resistance in series is always larger than the largest resistor.
2. The same current flows through both.

Answer: (a) 2 \( \Omega \)

Explanation:
1. 1/R = 1/6 + 1/3 = 3/6 = 1/2.
2. R = 2 \( \Omega \).

Teacher's Note:
1. The equivalent resistance in parallel is less than the smallest resistor.
2. For two resistors, R = r\( _1 \)r\( _2 \)/(r\( _1 \) + r\( _2 \)).

Answer: (a) the same current flows through each and the potential differences add up

Explanation:
1. Only one path exists, so the current is the same everywhere.
2. The total p.d. is the sum of the p.d.s across the individual resistors.

Teacher's Note:
1. In parallel, the p.d. is the same and the currents add.
2. Household appliances are connected in parallel.

Answer: (a) the potential difference is the same across each and the currents add up

Explanation:
1. All resistors are connected between the same two points, so each has the same p.d.
2. The main current divides among the branches and is equal to the sum of the branch currents.

Teacher's Note:
1. The lowest-resistance branch carries the largest current.
2. Adding a branch reduces the equivalent resistance.

Answer: (b) 0.5 A

Explanation:
1. R = 2 + 3 + 5 = 10 \( \Omega \).
2. I = V/R = 5/10 = 0.5 A.

Teacher's Note:
1. The current is the same in every resistor.
2. The p.d. across the 5 \( \Omega \) resistor is 2.5 V.

Answer: (d) 3 \( \Omega \)

Explanation:
1. R = (12 \( \times \) 4)/(12 + 4) = 48/16.
2. R = 3 \( \Omega \).

Teacher's Note:
1. The result is smaller than 4 \( \Omega \), as expected.
2. Check by 1/12 + 1/4 = 1/3.

Answer: (c) R/3

Explanation:
1. 1/R\( _{eq} \) = 1/R + 1/R + 1/R = 3/R.
2. R\( _{eq} \) = R/3.

Teacher's Note:
1. In series the equivalent resistance would be 3R.
2. Parallel combinations always reduce resistance.

Answer: (b) doubles

Explanation:
1. R = \( \rho \)l/A, so R is directly proportional to length for the same area.
2. Doubling l doubles R.

Teacher's Note:
1. Doubling the area of cross-section halves R.
2. Resistivity depends on the material and temperature.

Answer: (a) 2 V

Explanation:
1. V = IR = 0.5 \( \times \) 4.
2. V = 2 V.

Teacher's Note:
1. V = IR is Ohm's law written in terms of V.
2. Always check the units.

Answer: (d) its resistance increases as it gets hotter

Explanation:
1. The tungsten filament becomes very hot, and the resistance of a metal rises with temperature.
2. So V/I is not constant, and the filament does not obey Ohm's law over a wide range.

Teacher's Note:
1. Nichrome wire at constant temperature obeys Ohm's law well.
2. A diode does not obey Ohm's law.

Answer: (a) ohm metre

Explanation:
1. \( \rho \) = RA/l, so the unit is \( \Omega \) \( \times \) m\( ^2 \)/m.
2. The unit is \( \Omega \) m.

Teacher's Note:
1. Resistivity is a property of the material, not of the wire's size.
2. Copper has a very low resistivity.

Answer: (b) 1200 J

Explanation:
1. Energy W = VIt = 6 \( \times \) 2 \( \times \) 100.
2. W = 1200 J.

Teacher's Note:
1. The same energy appears as heat in the coil.
2. Power = VI = 12 W.

Answer: (b) the electrical work done divided by the heat produced in calories

Explanation:
1. Work done = VIt joules by the current; heat produced = (mass of water + water equivalent of the calorimeter) \( \times \) rise in temperature calories.
2. J = VIt / heat, about 4.2 J per calorie.

Teacher's Note:
1. J is expressed in joules per calorie.
2. The calorimeter should be well insulated.