Chapter 26: Problem 20
Two light bulbs for use at \(110 \mathrm{~V}\) are rated at \(60 \mathrm{~W}\) and \(100 \mathrm{~W}\), respectively. Which has the filament with lower resistance?
Chapter 26: Problem 20
Two light bulbs for use at \(110 \mathrm{~V}\) are rated at \(60 \mathrm{~W}\) and \(100 \mathrm{~W}\), respectively. Which has the filament with lower resistance?
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Get started for freeHow can you light a \(1.0-\mathrm{W}, 1.5-\mathrm{V}\) bulb with your \(12.0-V\) car battery?
Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?
How long would it take, in multiples of the time constant, \(\tau,\) for the capacitor in an \(\mathrm{RC}\) circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)
How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?
A capacitor \(C\) is initially uncharged. At time \(t=0,\) the capacitor is attached through a resistor \(R\) to a battery. The energy stored in the capacitor increases, eventually reaching a value \(U\) as \(t \rightarrow \infty\), After a time equal to the time constant \(\tau=R C\), the energy stored in the capacitor is given by a) \(U / e\). c) \(U(1-1 / e)^{2}\) b) \(U / e^{2}\) d) \(U(1-1 / e)\).
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