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a.
The current drawn when the bulb is lit.
a.
Answer to Problem 96A
3 A
Explanation of Solution
Given:
The resistance of the bulb when it is not lit, Rn=10.0 Ω
The resistance of the bulb when it is lit, Rl=40.0 Ω
The voltage applied, V=120 V
Formula used:
Ohm’s law,
V=IR
Where, I is the current, V is the potential and R is the resistance.
Calculation:
Using the above formula and the value of the resistance of the bulb when it is lit, the current drawn when the bulb is lit will be,
V=IlRl120=Il(40)Il=3 A
Conclusion:
Hence, the current drawn when the bulb is lit is 3 A .
b.
The current drawn at the instant when the bulb is turned on.
b.
Answer to Problem 96A
12 A
Explanation of Solution
Given:
The resistance of the bulb when it is not lit, Rn=10.0 Ω
The resistance of the bulb when it is lit, Rl=40.0 Ω
The voltage applied, V=120 V
Formula used:
Ohm’s law,
V=IR
Where, I is the current, V is the potential and R is the resistance.
Calculation:
When the bulb is turned on, the resistance of the bulb is the value of the resistance of the bulb when it is not lit.
Using the above formula and given values, the current drawn when the bulb turned on will be,
V=InRn120=In(10)In=12 A
Conclusion:
Hence, the current drawn when the bulb is turned on is 12 A .
c.
The case when the bulb used the maximum power.
c.
Answer to Problem 96A
When it is turned on
Explanation of Solution
Introduction:
The electric power (P) across a circuit is given by,
P=IV
Where, I is the current andV is the potential.
As can be seen from above formula, the power consumed is directly proportional to the current through the component. And,
From part (a), the current drawn when the bulb is lit, Il=3 A
From part (b), the current drawn when the bulb is turned on, In=12 A
Conclusion:
Hence, the bulb will use more power when it is turned on.
Chapter 22 Solutions
Glencoe Physics: Principles and Problems, Student Edition
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