The Electric Spark – Maharashtra Board Class 10 Solutions for Science and Technology (English Medium)
AlgebraGeometryScience & TechnologyHistoryGeography & EconomicsEnglish
Solution 1.1:
(b) 10^{3} A
Solution 1.2:
(a) series
Solution 1.3:
(b) 3.6 × 10^{6} Joule
Solution 1.4:
(b) 4 A
Solution 2:
 False. The SI unit of charge is coulomb.
 False. Voltmeter is always connected in parallel with the device.
 True
 True
 False. Resistivity of pure metal is less than alloys.
 False. The electric bulb consists of the filament whose melting point is high.
Solution 3:
Solution 4.1:
 A fuse is connected in series in the circuit to protect the appliances from damage due to high current.
 When the current in the circuit passes through the fuse, its temperature increases. When the current exceeds the specified value, the fuse should melt to break the circuit. Therefore, the material used for the fuse has a low melting point.
Solution 4.2:
 For the conduction of electricity through any material, two conditions are necessary. The first condition is potential difference, and the second is free charge carriers which can be accelerated by potential difference.
 Wood and glass do not have such charge carriers, so they cannot conduct electricity. Hence, wood and glass are good insulators.
Solution 4.3:
 We get light from the bulb due to heating of the filament by passing a current through it.
 The light emitted by the filament depends on the temperature of it. The intensity of light increases with the temperature.
 The melting point of a filament of a bulb is very high, so that it can withstand high temperature without melting and can give more light.
Solution 4.4:
 For the conduction of electricity, good conducting material is required.
 Copper and aluminum are good conductors as they provide a conducting path for motion of electrons through the circuit.
 They have free electrons which can be accelerated by a potential difference. Therefore, connecting wires in a circuit are made of copper and aluminum.
Solution 5.1:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another.
Solution 5.2:
If one coulomb of charge is passing through any cross section of a conductor in one second, the amount of current flowing through it is called one ampere.
1 A = 1 C/s
Solution 5.3:
If one ampere current flows through a conductor when 1 volt potential difference is applied across it, its resistance is one ohm.
Solution 5.4:
The electric level at a point is called the electric potential at that point.
Solution 5.5:
The resistivity of a conductor is the resistance of a conductor of unit length and unit area of cross section.
Solution 5.6:
Electric power is the electric work done per unit time. Its SI unit is watt (W).
Electric power (P) = _{}
Solution 6.1:
Solution 6.2:
Solution 6.3:
Solution 6.4:
Solution 7.1:
The electric current flowing in a metallic conductor is directly proportional to the potential difference across its terminals, provided the physical conditions of the conductor such as its length, area of cross section, temperature and material remain constant.
Solution 7.2:
The quantity of heat (H) generated in a conductor of resistance (R) when a current (I) flows through it for a time (t) is directly proportional to
 Square of the current
 Resistance of the conductor
 Time for which the current flows
Solution 8(a):

 Expression for the resistors in series:
R_{1}, R_{2} and R_{3} are the three resistors connected in series between points C and D as shown in the figure above. I is the current and V is the PD across points C and D.
Then R_{s} is the effective resistance in a circuit and V_{1}, V_{2} and V_{3 } are the potentials across the three resistors such that
V = V_{1 }+ V_{2} + V_{3} — (1)
Using Ohm’s law, the total potential difference
V = I R_{s
}and V_{1 }= IR_{1}, V_{2} = IR_{2} and V_{3} = IR_{3
}Substituting these values in equation 1,
we get
IR_{s} = IR_{1} + IR_{2} + IR_{3
} ∴ R_{s} = R_{1} + R_{2} + R_{3
}For n number of resistors connected in series,
_{ } Rs = R_{1} + R_{2} + R_{3} + R_{4} + —– + R_{n}
_{ }2. Expression for the resistors in parallel:
R_{1},R_{2} and R_{3 }are the three resistors connected in parallel between points C and D as shown in the figure.
Let I_{1},I_{2 }and I_{3} be the currents passing through R_{1, }R_{2} and R_{3}, respectively. Let V be the potential difference between points C and D. The current I will be given by
I = I_{1 }+ I_{2} + I_{3} —– (1)
Let R_{p} be the effective resistance in the circuit.
Using Ohm’s law, we get
I_{1} = _{}
Putting these values in equation 1, we get
Solution 8(b):
The resistance R of a conductor is directly proportional to its length l and inversely proportional to its area of cross section A.
Solution 9.1:
 Wires P and Q have the same resistivity as they are made of the same metal.
 Wires P and Q have different resistances.
_{Resistance } _{}
where l is length and A is the area of cross section of a conductor. Here r and l are the same for P and Q, but A is not. Hence, the wires have different resistances.
Solution 9.2:
Silver will be the best conductor, because when the resistivity is lower, the conductivity is greater.
Solution 9.3:
Solution 9.4:
Assuming that the bulbs have a filament of the same length and the same area of cross section but are made of metals with different resistivity, then the bulb with higher resistance will be brighter.
Solution 10.1(a):
The resistors of resistances 20 Ω, 3.5 Ω and 4.9 Ω are connected in series. Therefore,
R_{4} = 20 + 3.5 + 4.9 = 28.4 Ω
Solution 10.1(b):
Parallel combination of resistors:
Solution 10.2:
Solution 10.3:
Solution 10.4:
Solution 10.5:
Solution 10.6:
Solution 10.7:
Solution 10.8:
Solution 10.9: