A series circuit consists of a battery of negligible internal resistance, a vari
ID: 1703586 • Letter: A
Question
A series circuit consists of a battery of negligible internal resistance, a variable resistor, and an electric motor of negligible resistance. The current in the circuit is 2 amperes when the resistance in the circuit is adjusted to 10 ohms. Under these conditions the motor lifts a 1-kilogram mass vertically at a constant speed of 2 meters per second.
a. Determine the electrical power that is
i. dissipated in the resistor
ii. used by the motor in lifting the mass
iii. supplied by the battery
b. Determine the potential difference across
i. the resistor
ii. the motor
iii. the battery
The resistor is now adjusted until the mass rises vertically at a constant speed of 3 meters per second. The voltage drop across the motor is proportional to the speed of the motor, and the current remains constant.
c. Determine the voltage drop across the motor
d. Determine the new resistance in the circuit
Explanation / Answer
(1)
Current I = 2 A
Resistance R = 10
Mass m = 1 Kg
Speed v = 2 m/s
(a)Power
i. dissipated in the resistor = i^2 R = 4 * 10 = 40 W
ii. used by the motor in lifting the mass = mg v = 1*9.8*2 = 19.6 W
iii. supplied by the battery = 40 + 19.6 = 59.6 W
(b) the potential difference across
i. the resistor = 40/2 = 20
ii. the motor = 19.6/2 = 9.8
iii. the battery = 59.6/2 = 29.8
(c)
New current I = 2 A
New resistance R' = ?
New speed v = 3 m/s
the voltage drop across the motor = 9.8 * (3/2)
= 14.7 V
(d)
the new resistance in the circuit R' = R + (14.7/3)
= 10 + 4.9
= 14.9
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