Test: Electricity - 2 - Class 10 MCQ

Correct relation is current I = 
⇒ q = It

6Ω is shorted
as current will pass from wire having no resistance
so effective resistance is 4Ω.

As resistivity of the material of a resistor of length l, cross-section area A and resistance R is given by :ρ = R.A/I, hence SI unit of resistivity is ohm -metre (Ω m).

Total power used, P - P₁ + P₂ = 1500 + 500 = 2000 W.
Current drawn from the supply,

If four resistors each of 1Ω are connected in parallel, the effective resistance will be 0.25Ω..

As, in parralel connection,

1/R_eff. = 1/R₁ + 1/R₂ + ....

• The fuse wire repeatedly gets burnt when used with a good heater because the current flowing through the circuit exceeds the capacity of the fuse wire. To prevent this, we need to increase the current-carrying capacity of the fuse wire.
• The current-carrying capacity of a fuse wire depends on its material, length, and cross-sectional area (which is related to its radius).
• A wire with a larger radius has a greater cross-sectional area, which allows it to carry more current before it heats up to the point of melting.

Resistivity of a given conductor depends only on its material and the temperature. Resistivity does not depend on the dimensions of the conductor.

Original power consumed, P = V²/R
When used in parallel 

R_P = R/4
∴ New power consumed when two halves in parallel

Value of series combination of 2Ω and 2Ω resistances = 4Ω. As it is now connected in parallel to 4Ω resistance, hence the effective resistance R between the points A and B will be

An alloy ‘nichrome’ is used to prepare the heating element of an electric iron because its resistivity is high, its melting point is high and it does not oxidise easily.

When the lamp is on use, it is hot. 
Using joules law of heating ; 
P =  V² / R 
R  =  220² /100  = 484 ohm. 
It is given resistance of hot lamp is 20 times the cold lamp. 
Hence, R cold  = 484/10 = 48.4 ohm.

Resistance R = V²/P where V = voltage/potential difference and P - power
∴ electrical resistance of 100 W lamp,
and electrical resistance of 40 W lamp,

Using power, 

For the same voltage, R ∝ 1/P
More the power, lesser the resistance.
Accordingly, R₂ < R₁

When a wire of 8Ω resistance is bent in the form of a closed circle, resistance of each half (semicircular) part is 8/2 Ω = 4Ω. 
As these two parts are connected in parallel across points A and B, hence effective resistance will be 2Ω. 

In series, R_s = R₁ + R₂ = 1OΩ
In parallel, 

R₁R₂=12/5
R₁+R₂
R₁R₂=12/5x(R₁+R₂)
R₁R₂=12/5x10
R₁R₂=24
so option (B) fits these two equation very well

A rheostat is used to increase or decrease the magnitude of the current flowing through a circuit without change in the voltage source.

In parallel, potential difference across each resistor will remain same. So, current through 10Ω resistor

I-V graph shown in figure (b) is correct because I ∝ 7.