A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
pacitor is zero. If a 24-volt battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at any time t.
Q(t)=?
Determine the charge on the capacitor at t = 0.001 s, and at t = 0.01 s. (If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.)
Q(t)=?
Determine the charge on the capacitor at t = 0.001 s, and at t = 0.01 s. (If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.)
1 answer:
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So an exponential equation is gonna be used for this problem. The formula for such is 
, in which the a variable is the initial value and the b variable is the growth/decay rate.
For this situation, 1200 is gonna be your a variable since you've started out with it. And as for your b variable, its gonna be 107.3%, or 1.073, since it's increasing.
Your equation should look like this:
and from here just plug in 10 into the x variable and solve.
Multiply (1.073)^10 (don't round answers until the end), and take that answer and multiply it with 1200, your answer should be $2427.61.
For this situation, 1200 is gonna be your a variable since you've started out with it. And as for your b variable, its gonna be 107.3%, or 1.073, since it's increasing.
Your equation should look like this:
Multiply (1.073)^10 (don't round answers until the end), and take that answer and multiply it with 1200, your answer should be $2427.61.