Asked by ann
A 79.4-Ω and a 47.7-Ω resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.296 A. When the 47.7-Ω resistor is disconnected, the current from the battery drops to 0.126 A.
(a) Determine the emf.
? V
(b) Determine the internal resistance of the battery.
? Ω
(a) Determine the emf.
? V
(b) Determine the internal resistance of the battery.
? Ω
Answers
Answered by
drwls
() The effective resistance R of the two resistors in parallel is given by
(1/R) = 1/79.4 +1/47.7 = 1/3.356*10^-2
R = 29.8 ohms
Let Ri be the internal battery resistance.
(a) V = 0.126*(79.4 + Ri)
(b) V = 0.296*(29.8 + Ri)
1 = 2.349*(29.8+R1)/(79.4 + Ri)
0.4257 = (29.8 + Ri)/(79.4 + Ri)
Solve for Ri; then use Ri in equation (a) to get V.
33.80 + 0.4257 R1 = 29.8 + Ri
0.5743 Ri = 4
Ri = 6.97 ohms
V = 10.88 Volts
(1/R) = 1/79.4 +1/47.7 = 1/3.356*10^-2
R = 29.8 ohms
Let Ri be the internal battery resistance.
(a) V = 0.126*(79.4 + Ri)
(b) V = 0.296*(29.8 + Ri)
1 = 2.349*(29.8+R1)/(79.4 + Ri)
0.4257 = (29.8 + Ri)/(79.4 + Ri)
Solve for Ri; then use Ri in equation (a) to get V.
33.80 + 0.4257 R1 = 29.8 + Ri
0.5743 Ri = 4
Ri = 6.97 ohms
V = 10.88 Volts
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