Science Olympiad Test: Electricity- 2 - Free MCQ with solutions Class 10

• A multimeter can measure voltage, current, and resistance.
• It can also be used to generate square wave and for checking shorts in a circuit.
• An ammeter is an instrument for measuring electric current in amperes while Voltmeter is an instrument for measuring electric potential in volts.
• An ohmmeter is an electrical instrument that measures electrical resistance.

The voltmeter should be connected parallel to the resistor and the ammeter should be in series with the resistor.

From Resistance,

Where R = resistance, l = length, A = area of cross-section and ρ = resistivity
If l = 1 m and A = 1 m², then
R = ρ i.e.

• Resistivity is numerically equal to the resistance of a substance having unit area of cross-section and unit length.
• Resistivity is the intrinsic property of the substance. It is independent of shape and size of the body i.e. on length and area. Hence, option 1 and 2 is incorrect.
• Formula of resistivity is 
  Where, m = mass of electron, n = Number of electron per unit volume of the conductor, τ = relaxation time.
• Therefore, the resistivity of a wire depends on its material (n)
• For different substances their resistivity is also different as ‘n’ will be different.
• Resistivity depends on the temperature.

By Ohm’s Law V = IR
⇒ R= V/I = slope of V-I graph 
Therefore, the slope of the V-I graph represents the effective resistance. 
Now in series combination of the resistance all the resistances get summed up 
R_eq = R₁ + R₂ + R_{3+ ………….+ Rn.} 
Therefore, the series combination will have the maximum resistance than the other resistors.
Hence the line with maximum slope (line C) will give the series combination of the other two resistors.

When length (l) of a wire doubled (2 l), its cross-sectional area (A) is reduced to half (A/2) as the volume of the wire is constant.
Now resistance of the wire

Q = I t = 0.5 × (2 × 60 min × 60 sec) = 3600 seconds
Note: Wattage of the bulb does not affect the flow of charge in this case

Resistance of each half = 1Ω/2 = 0.5 Ω (R ∝ l) Since these two halves are in parallel, Resultant resistance = 

The charge on one electron, e = 1.6 × 10^–19 C
Number of electrons constituting 1 coulomb is given by n = Q/e

V = W/Q = W/ I t and V = I R (work done is equal to heat generated)
⇒ W = VI W = I² R t

V = I R = 5 A × 4 Ω = 20 V

Three resistors of 12 Ω each in parallel have effective resistance R_e, as

This effective R_e is in series with two other resistors of 2 Ω and 4 Ω. Thus, overall resistance of whole circuit is
R_c = R_e + R₁ + R₂ = 4 + 2 + 4 = 10 Ω

W = QV = 5 × 10 = 50 J 

Consider the wire of resistance 'R' is cut into five equal parts
Thus, Resistance of each part will be R/5
Equivalent resistance R' will be given by,
1/R' = 5/R + 5/R + 5/R + 5/R + 5/R  ...( resistances connected in parallel)
Thus, equivalent resistance R' = R/25.

The current flowing in a conductor is halved if the resistance of the conductor is doubled keeping the potential difference across it the same.
By ohm's law,
Voltage, V = current(I) * resistance (R)
V = IR
I = V/R
when voltage doubled,
new voltage is V(2)
V(2)=2V
V(2) = I(2) * R
substituting V(2) as 2V,
2V = I(2) * R
hence new current, I(2)
I(2) = 2V/R
but V/R = I
Hence, I(2) = 2V/I= 2I.
Hence current also doubles.

Ohm's law: The current flowing through a conductor is directly proportional to the voltage applied across the conductor means I ∝ V.