1. The circuit in the drawing shows two resistors, a C = 8.5 µF capacitor, and a
ID: 1267569 • Letter: 1
Question
1.
The circuit in the drawing shows two resistors, a C = 8.5 µF capacitor, and a V = 9 V battery. When the capacitor is fully charged, what is the magnitude q of the charge on one of its plates?
2.
A 71.0-? and a 44.1-? resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.266 A. When the 44.1-? resistor is disconnected, the current from the battery drops to 0.110 A.
(a) Determine the emf.
V
(b) Determine the internal resistance of the battery.
?
Explanation / Answer
When 71 ohm and 44.1 ohm are in parallel, then their resultant
= 71 * 44.1/(71 + 44.1) = 27.20 ohm
Let r = internal resistance of the battery
Then total resistance = 27.20 + r
Current = 0.266 A
Let E = emf
E = (27.20 + r) * 0.266 -----------------------(1)
When 44.1 ohm is removed, then 71 ohm and r are in series.
Therefore, total resistance = r + 71
Current = 0.110 A
Therefore E = (r+71) * 0.110 --------------------(2)
From (1) and (2),
(27.20 + r) * 0.266 = (r+71) * 0.110
Or 7.236 + 0.266 r = 0.110 r + 7.81
Or 0.156 r = 0.574
Or r = 0.574/0.156 = 3.68 ohm
Putting this value of r in (2),
E = (3.68 + 71) * 0.110
= 8.21 V
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.