A capacitor of capacity 4F is charged to SOV andanother capacitor of c

1.	A capacitor of capacity 4F is charged to SOV andanother capacitor of capacity 6F is charged to30V. When they are connected. then the energylost by Apulf capacitori(1) 7.8 m(2) 46 m() 3.2 ml(4) 2.5 m
Answer» EXPLANATION: HEAT loss will be, $heat = \frac{1}{2} \frac{c1 \times c2}{c1 + c2} {(v1 - v2)}^{2}$ Here, c1 =4F c2= 6F V1= 50V V2= 30V Solving, we will GET heat loss=480 JOULE

A capacitor of capacity 4F is charged to SOV andanother capacitor of capacity 6F is charged to30V. When they are connected. then the energylost by Apulf capacitori(1) 7.8 m(2) 46 m() 3.2 ml(4) 2.5 m

Answer»

EXPLANATION:

HEAT loss will be,

$heat = \frac{1}{2} \frac{c1 \times c2}{c1 + c2} {(v1 - v2)}^{2}$

Here, c1 =4F

c2= 6F

V1= 50V

V2= 30V

Solving, we will GET heat loss=480 JOULE

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A capacitor of capacity 4F is charged to SOV andanother capacitor of capacity 6F is charged to30V. When they are connected. then the energylost by Apulf capacitori(1) 7.8 m(2) 46 m() 3.2 ml(4) 2.5 m​

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A capacitor of capacity 4F is charged to SOV andanother capacitor of capacity 6F is charged to30V. When they are connected. then the energylost by Apulf capacitori(1) 7.8 m(2) 46 m() 3.2 ml(4) 2.5 m