The electromotive force of the voltage supply is 120V and the internal resistanc

Question

The electromotive force of the voltage supply is 120V and the internal resistance is 5W.

R(bulb)= 12ohms

a) In order for the bulb to glow appreciably, the current flow through the filament must be at least 0.7 A.  What is the maximum resistance the resistor can have for the bulb to glow?

b) When the circuit is attached, the actual current through the bulb is 1.0 A.  What is the rate of resistive heat loss through the resistor?

c) Replacing the bulb with a completely discharged 0.05F capacitor, we then send electrical current through the resistor to charge the capacitor.  How long will it take to charge the capacitor to 90% capacity

Explanation / Answer

Part A)

V = IR

120 = (.7)(5 + 12 + R)

R = 154 Ohms

Part B)

The real R

120 = 1(5 + 12 + R)

R = 103

P = V^2/R

P = (120^2)/103

P = 140 Watts

Part C)
Apply q = Q(1 - e^-t/RC)

R = 103 + 5 = 108 Ohms

.9 = (1 - e^-t/(108)(.05))

t = 12.4 sec

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