A 12.0-μ F capacitor is charged to a potential of 50.0 V and then discharged thr

A 12.0-μ F capacitor is charged to a...

Key Concepts

Half-life in Exponential Decay

The half-life of a decaying quantity in an exponential process is the time required for the quantity, such as charge or energy, to reduce to half its initial value. In an RC circuit, calculating the half-life involves solving the exponential decay equation for the time at which the quantity equals one half, which is directly related to the time constant.

Capacitor Stored Energy

The energy stored in a capacitor is given by the expression (1/2)CV², where C is the capacitance and V is the voltage across the capacitor. As the capacitor discharges, the energy in the capacitor decreases not only due to the drop in charge but also because the voltage decreases, affecting the energy quadratically.

RC Time Constant

The RC time constant, denoted by ? (tau), is the product of the resistance (R) and the capacitance (C) in an RC circuit. It characterizes the rate at which the capacitor discharges its stored charge through the resistor. A larger time constant indicates a slower discharge rate, while a smaller one means a faster discharge.

Exponential Decay in RC Circuits

When a charged capacitor discharges through a resistor, the voltage, charge, and consequently the energy stored decay exponentially over time. This decay is mathematically described by exponential functions, where the value at any time is a fixed fraction of the initial value, determined by the time constant.

Transcript

00:01 In the given problem, a capacitor having the capacitance value of 12 micro ferret is charged with the help of a source of emf having the voltage is equal to 50 .0 volt.

00:18 Then it is discharged through a resistance having a value of 175 oom.

00:27 Now in the first part of the problem we have to find the time taken by this capacitor to lose half of its charge.

00:43 So to find it the expression for discharging circuit for discharging circuit, the expression for instantaneous charge remaining over the capacitor plates is given as qt means charge as a function of time is equal to q0 multiplied by e ratio of power minus t by rc where this rc is the time constant and q0 is the maximum charge stored over the plates of capacitor and this maximum charge is given as the product of c and v.

01:53 But we need not to use it because the final charge remaining over the plates of the capacitor has been given as a fraction of the total maximum charge.

02:06 Means to say this qt has been given as q0 by 2.

02:11 So using this value q0 by 2 in this given equation, it becomes q0 by 2 is equal to q0 into e ratio the power minus t by rc and we have to find the time taken the charge q0 to drop up to the value of q0 by 2 this q0 get cancelled so the equation will become 0 .5 is equal to a ratio bar minus t by the product of r and and c here r is 175 om multiplied multiplied by the capacitance which is 12 into 10 which is 4 minus 6 so finally this is 0 .5 is equal to e rach to the power minus t divided by 2 .1 into 10 dash the power minus 3 so if we take natural law of both the sides it becomes l n for natural logarithm of 0 .5 is equal to minus t by 2 .1 into 10 dash par minus 3.

03:27 So the expression for time here comes out to be t is equal to minus 2 .1 into 10 dash bar minus 3 multiplied by natural logarithm of 0 .5 and this gives us the final answer for the time as 1 .46 into 10 dash bar minus 3 seconds or we can say the time is 1 .46 milliseconds and it becomes the answer for the first part of the problem.

04:06 Now in the second part of the problem we have to find the time in which its energy, the energy stored over the capacitor will be reduced.

04:16 Reduced to half of its initial energy...

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