Two capacitors, 3microfaraday and 4microfaraday are individually charged across a 6 volt battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the common potential?
(a) 6 volt (b)6/7 volt (c) 2 volt
(d) 3/2 volt
Answers
Answered by
3
d= volume of the oil drop/ area of the film
= 4/3 *pi* (0.025cm)^3/(pi* (10 cm)^2
= 2.08X10^-7 cm
= 4/3 *pi* (0.025cm)^3/(pi* (10 cm)^2
= 2.08X10^-7 cm
Answered by
3
Common potential according to given data is (b)6/7 volt.
Charge in capacitor is calculated by the formula - Q=C*V
Q is charge, C is capacitance and V is voltage.
For 3 microfaraday and 4 microfaraday individual capacitors -
Q= 3*6 = 12 microcoulomb and
Q= 4*6 = 24 microcoulomb.
For connected unit, common potential or V will be calculated by the same formula -
V= Q/C.
Since, negative plate is attached to positive plate, thus Q will be-
Q= 24-18 =6 microcoulomb.
C= 3+4 =7 microfaraday.
V= 6/7 V.
Thus, common potential will be 6/7 V.
Similar questions