Question

a)
A parallel plate capacitor of 2 uF at 150V at a
parallel plate capacitor of 1 uF at 300 V are
connected in parallel. The total energy lost by the
capacitors is equal to​

Answer 1

Answer:

sjdkkdmdmd d bd ejsjjxb zjdnf xj rkzbebe r. f xbdjrjdn. d f. f f. dkfkkrifof

Explanation:

jsjdjdjdjdjdjheueudirjgsac. didorokdkekrne r dkf. f t t dbskksidudiskek

Answer 2

Given:

C₁ = 2μF, V₁ = 150V

C₂ = 1μF, V₂ = 300V

To find:

Total energy loss

Solution:

First capacitor capacitance, C₁ = 2μF

Potential of first capacitor, V₁ = 150V

Second capacitor capacitance, C₂ = 1μF

Potential of second capacitor, V₂ = 300V

they are connected parallelly. So, common potential can be calculated as-

$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

$V = \frac{2 \times 150 + 1 \times 300}{2 + 1}$

V = 200V

So, final charge on capacitors are

q₁' = C₁V = 2× 200 = 400μC

q₂' = C₂V = 1 × 200 = 200μC

so, total loss in energy is,

$\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$

$\Delta U = \frac{1}{2} \frac{2 \times 1}{(2 + 1)} (150 - 300)^2$× 10⁻⁶

ΔU = 7500μJ

Therefore the energy loss is 7500μC.