I have Three questions, Please show all corresponding work for future reference.

Question

I have Three questions, Please show all corresponding work for future reference.

1. A 38.42 microF capacitor initially charged to 37.14 microC is discharged through a 1.45 kilo Ohm resistor. How long, in ms, does it take to reduce the capacitor's charge to 7.03 micro C?

2. A 33.65 g horizontal metal bar, 12.35 cm long, is free to slide up and down between two tall, vertical metal rods. A 0.16 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a 2.19 ohm resistor is connected across the two rods at the top. Then the bar is dropped. What is the terminal speed at which the bar falls? Assume that the bar remains horizontal and in contact with the rods at all times.

3. A real battery is not just an emf. We can model a real 1.98 V battery as a 1.98 V emf in series with a resistor known as the "internal resistance", as shown in the figure. A typical battery has 1.0 ohm internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.98 V , the value of the emf. Suppose the terminals of this battery are connected to a 2.53 ohm resistor.What is the potential difference between the terminals of the battery?

Explanation / Answer

1.

capacitor, C = 38.42 microF
initial charge Qo= 37.14 microC
final charge Qf= 7.03 micro C
resistor,R = 1.45 kilo Ohm

time constant = RC = 38.42 microF * 1.45 kilo Ohm = 0.0557
Qf = Qo e^(-t/RC)
7.03 = 37.14 e^(-t/0.0557)
e^(-t/0.0557) = 7.03 / 37.14 = 0.1892
taking log on both sides
-t/0.0557 = ln 0.1892 = -1.6645
t = 0.0927 s = 92.7 milli sec

3.

Battery Voltage Vs= 1.98 V
internal resistance,r = 1 ohm
external resistor R = 2.53 ohm

when external resistor is connected to battery . we have the following kvl equation :
Vs = i(r+R)
98 = i(1+2.53) = i 3.53
i = 0.56 amps
potential difference between the terminals of the battery = 1.98 - i*r = 1.98 - 0.56*1 = 1.42 volts

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