Test: Electrical Energy and Power (December 4) - NEET MCQ

Resistance of a wire, R = ρl/A
Rate of energy dissipated as heat is
H = V²/R = V²A/ρl
For a wire 1,

For a wire 2,

= 
For a wire 3,

For a wire 4,  

= 
∴H₃ > H₁ > H₂ > H₄.

Power, P = V²/R
As the resistance of the bulb is constant
∴ ΔP/P = 2ΔV / V
% decrease in power = ΔP/P × 100
= 2ΔV/V × 100
= 2 × 2.5% = 5%

The resistance of heater coil,

=  = 400Ω
The resistance of either half part = 200 Ω.
Equivalent resistance when both parts are connected in parallel.

The energy liberated per second when the combination is connected to a source of 200 V,
=  
= 400 J

Given that

As, Resistor (R) = 10 ohm
Current (I) = 1.5 A
Time (T) = 1 minute = 60 seconds

We know that
Power = I²R

Hence,
Energy E = I²R⋅t
E = 1.5² × 10 × 60
E = 2.25 × 600
E = 1350 J

P = 1350 / 60 = 22.5 W

Therefore, the energy dissipated in the resistor in 1 minute will be 1350 Joules or 22.5 Watt.

In a circuit X, both the resistance are in parallel, Therefore V is the same and, Power (energy transferred in unit time) is
V²/2 = W
In a circuit Y, both resistance are in series.
Therefore, V_B + V_B' = V or V_B = V/2
In a circuit Y, power supplied to 
B  = 

⇒ R_B = 25Ω (Q Both were in parallel, same V)

∴ Heat in 5 min = 

= 100kJ

Q = at − bt² ⇒ 

Current becomes zero at time t = a/2b

Total heat produced = 

Current through 2Ω,

Hence produced per second, H₁
= 
Current through, 4Ω, 

Hence produced per second H₂
= 
Current through, 3Ω,I
Heat produced, H₃ = I² × 3 = 3I²
= 
∴ H₁ : H₂ : H₃ = 8 : 4 : 27

For this circuit → power generated in x is maximum when x = r.
In given circuit is

For power to be minimum in y,  y = 2 + 6x / 6 + x = 4
⇒ x = 3Ω

The resistance of the bulb rated 100V, 1000W is

Hence, when the bulb is connected to 25 volt source the output power is

= 62.5 Watts