Four capacitors with capacitances C_1 = 1,&mu | vedclass

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Question

$1\left(\mathrm{V}_{\mathrm{A}}-3\right)+\frac{3}{2} \mathrm{V}_{\mathrm{A}}=0$

$\mathrm{V}_{\mathrm{A}}=12$

$\left(\mathrm{V}_{\mathrm{B}}-30\r

Vedclass · Accepted Answer

$13$

Vedclass · Answer

$1\left(\mathrm{V}_{\mathrm{A}}-3\right)+\frac{3}{2} \mathrm{V}_{\mathrm{A}}=0$

$\mathrm{V}_{\mathrm{A}}=12$

$\left(\mathrm{V}_{\mathrm{B}}-30\right) \frac{5}{2}+\frac{\mathrm{V}_{\mathrm{B}}}{2}=0$

$\mathrm{V}_{\mathrm{B}}=25$

$\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=13 \mathrm{\,Volt}.$

Four capacitors with capacitances $C_1 = 1\,μF, C_2 = 1.5\, μF, C_3 = 2.5\, μF$ and $C_4 = 0.5\, μF$ are connected as shown and are connected to a $30\, volt$ source. The potential difference between points $B$ and $A$ is....$V$

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