# AP Physics?

Two resistors A and B are connected in parallel across a 12V battery. The current through B is found to be 1.1A. When the two resistors

are connected in series to the 12 V battery,

a voltmeter connected across resistor A measures a voltage of 3.2V.

Find the resistance of B.

Answer in units of Ω.

Find the resistance of A.

Answer in units of Ω.

### 7 Answers

- PhilomelLv 71 month ago
12v/1.1A=10.9Ω.

12v-3.2v=8.8v.

I=8.8/10.9Ω=.807A

R=E/I=3.2/.807=3.96Ω

RA & RB=3.96Ω & 10.9Ω.

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- oubaasLv 71 month ago
in parallel

Rb = 12/1.1 ohm

in series

Va = 3.2 V

Vb = 12-Va = 8.8 V

Ra = Rb*Va/Vb = 12/1.1*3.2/8.8 = 4.0 ohm

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- oldschoolLv 71 month ago
Ra = 12V/1.1A = 120/11 Ω

When Ra and Rb are in series across 12V and Ra has 3.2V across it, Rb must have 12-3.2 = 8.8V across it. The current through both Ra and Rb: i = 8.8/(120/11) = 242/300 or about 0.807A. Therefore Ra = 3.2V/0.807A = 960/242 Ω or about 3.97Ω

Rb = 3.97Ω or about 4.0Ω <-----

Ra = 120/11 or about 10.9Ω <-----

Check 12/(3.97+10.9) = 0.807A Checks

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- oldprofLv 71 month ago
Rb = Vb/Ib = 12/1.1 = 10.9 ohms. ANS.

V = Va + Vb = 12 = 3.2 + Vb; so that Vb = 8.8 v = IRb so that I = Vb/Rb and Va = IRa so Ra = Va/I = 3.2/(Vb/Rb) = 3.2/(8.8/10.9) = 3.96 ohms. ANS

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- billrussell42Lv 71 month ago
I typed in the answer, but Yahoo had problems.

rather than retype it all, look at the screen shot below, which I managed to salvage.

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- KnrLv 41 month ago
for parallel combination i1R1= i2R2. is., i2= 1.1xR1/R2

for series combination R1= 3.2/I and 8.8 /I, so that R1/R2= 32/88

then solving these two eqns we can get R1=5.87 ohm and R2 = 2.13 ohm.

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- Anonymous1 month ago
V = I * R. resistance of B = 12 V / 1.1 amp

When you connect them in series, the total voltage drop across them combined is still 12V. Since the voltage drop across A is 3.2 V, the voltage drop b is 12V - 3.2V = 8.8 V. You know the resistance from above so you can calculate the current flow or 0.807 amps. This is also the current flow through A since they are in series. V = I * R, R = 3.97 ohms.

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