Answer:
Explanation:
As we know that when battery is in open circuit then the potential difference of the cell is known as EMF
so EMF is given as
now when the battery is connected across a resistance of 200 ohm then there is current flowing through the battery
it is given as
Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
(1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC