i. An electric current in conductors is due to the flow of




... Answer is (d) free electrons
A metallic conductor has a large number of free electrons in it. These free electrons begin to drift from the low potential to the high potential region when a potential difference is applied across the ends of a metallic wire. Thus, the Current in a conductor is due to the motion of the free electrons.


ii. What is the voltage across a 6 Ω resistor when 3A of current passes through it?




... Answer is (c) 18 V
Given: Resistance (R) = 6 Ω Current ( I ) = 3 A To Find: Voltage V = ? Solution: V = I x R = 3 A x 6 Ω = 18 volts


iii. What happens to the intensity or the brightness of the lamps connected in series as more and more lamps are added?




... Answer is (b) decreases
 The networks of Resistances connected in Series can also be thought of as “voltage dividers”. So in this case the more the bulb is connected in series divides the voltage and thus brightness or intensity of the bulb decrease.


iv. Why should household appliances be connected in parallel with the voltage source?




... Answer is (c) to provide each appliance the same voltage as the power source
The advantage of home appliances connected in a parallel network is that the Voltage across each appliance remains the same i.e the voltage does not drop across each appliance


v. Electric potential and e.m.f




... Answer is (a) are the same terms
Both are the same term having the same unit as Volt. The emf (electromotive force) is the potential difference between the terminals of a battery when the circuit is open i.e. no current is flowing. The potential difference is the voltage across the terminals of the battery when the current is being drawn from it to an external i.e. a Closed circuit.


vi. When we double the voltage in a simple electric circuit, we double the




... Answer is (a) current
Ohm's law formula V = IR ⇒ V ∝ I Since Resistance does not change with a change in voltage. The change occurs only in the current. 2V makes 2I


vii. If we double both the current and the voltage in a circuit while keeping its resistance constant, the power: :




... Answer is (d) Quadruple
Formulas of Power, P = VI , P = I²R, P = V²/ R Here we will use P=VI Replacing V with 2V and I with 2I ⇒ P= 2Vx 2I ⇒ P = 4 VI. Hence the power becomes four times


viii. What is the power rating of a lamp connected to a 12 V source when it carries a 2.5A?




... Answer is (c) 30 W
Given : V = 12 V Current = 2.5 A To Find : Power of Lamp P = ? Solution: Her the formula for Power P = VI ⇒ P = 12 × 2.5 ⇒ P = 30 watt


ix.. The combined resistance of two identical resistors, connected in series is 8Ω. Their combined resistance in a parallel arrangement will be:




... Answer is (a) 2 Ω
Given: Req (for Series network) = 8 Ω . To Find: First, we will find the value of each resister R1=R2=R=? then Req (for parallel network) = ? Solution: Formula of Series network Req (Series) = R1 + R2 (as R1=R2=R identical resisters) ⇒ Req (Series) = R + R = 2R ⇒ R = Req (Series) / 2 = 8 /2 = 4 Ω ⇒ Now for Parallel network 1 / Req = 1/R1 + 1/R2 ⇒ 1 / Req = 1/R + 1/R (as R1=R2=R identical resisters) ⇒ 1 / Req = 2 /R ⇒ 1 / Req = 2 / 4 (putting R=4 Ω) ⇒ 1 / Req = 1/2 ⇒ Req (for Parallel resisters network) = 2 Ω
 
 

VIP Note: If you have Learned the above MCQs, then examine yourself by taking this Simple MCQS Quiz (Click here to play the Quiz)


 
Unit 14: Current Electricity 
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List of Class:10 MCQs Quizzes

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