Question

Two identical capacitors store different amounts of energy: capacitor A stores 2.0 x 10-3 J, and capacitor B stores 1.8 x 10-4 J. The voltage across the plates of capacitor B is 12 V. Find the voltage across the plates of capacitor A.

Answer 1

Answer:

39.9 V

Explanation:

Given that

Capacitor are indentical that is why the value of C will be same for both the capacitor.

We know that energy stored in the capacitor is given as

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

V=Voltage

C=Capacitance

For A :

E₁= 2 x 10⁻³ J

Voltage = V₁

For B :

E₂=1.8 x 10⁻⁴ J

V₂ = 12 V

$E_1=\dfrac{1}{2}\times CV_1^2$

$2\times 10^{-3}\times 2=CV_1^2$

$E_2=\dfrac{1}{2}\times CV_2^2$

$1.8\times 10^{-4}\times 2=C\times 12^2$

Now by using above two equation

$\dfrac{2\times 10^{-3}\times 2}{1.8\times 10^{-4}\times 2}=\dfrac{CV_1^2}{C\times 12^2}\\11.11\times 12^2=V_1^2\\V_1=\sqrt{1599.84}\\V_1=39.9\ V$

Therefore the voltage on capacitor A will be 39.9 V