Question

Two capacitors with capacitances of 1.5

Answer 1

Answer:

$E = 1.9 * 10^{-3}J$

Explanation:

The complete question is summarized as thus:

Given

$C_1 =1.5\mu F$

$C_2 =0.25\mu F$

$V = 50V$ --- voltage

Connect type: Parallel

Required

The potential energy stored in the $1.5\mu F$ capacitor

The energy stored is calculated as:

$E = \frac{1}{2}CV^2$

Where:

$C = C_1 =1.5\mu F$ --- the capacitor

$V = 50V$ --- the voltage across the capacitor

So, we have:

$E = \frac{1}{2} * 1.5\mu F * (50V)^2$

Convert to Farad

$E = \frac{1}{2} * 1.5 * 10^{-6} F * (50V)^2$

$E = \frac{1}{2} * 1.5 * 10^{-6} F * 2500V^2$

$E = 0.001875J$

Rewrite as:

$E = 1.875 * 10^{-3}J$

Approximate

$E = 1.9 * 10^{-3}J$