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2 years ago

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7) Three resistors having resistances of 4.0 Ω, 6.0 Ω, and 10.0 Ω are connected in parallel. If the combination is connected in

series with an ideal 12-V battery and a 2.0-Ω resistor, what is the current through the 10.0-Ω resistor?
A) 0.59 A

B) 2.7 A

C) 6.4 A

D) 11.2 A

E) 16 A

E) 16 A

1 answer:

viva [34]2 years ago

6 0

Answer:

A, 0.59A

Explanation:

The total resistance in the circuit is the resistances in parallel plus that in series.

Total resistance for those in parallel is;

1/(1/4 +1/6 +1/10) = 1/ (15+10+6 /60)

1/(31/60)= 60/31 ohms

Hence total resistance of the circuit is;

60/31 + 2 = (60+62)/31 = 122/31=3.94 ohms

To calculate the current flowing through the 10ohm resistance we need to know the voltage drop by subtracting the voltage drop in the 2ohm resistance from the total voltage drop.

Voltage drop on the 2 ohm resistance is;

Current on the 2 ohm resistor × 2 ohms

V = I ×R ; I - current

R - resistance

Current drop on the 2ohm resistance is;

Total voltage in the circuit/ total resistance in the circuit

12/3.94= 3.05A

Voltage drop on the 2 ohm resistance;

3.05 × 2 = 6.10volts

Hence voltage drop on the parallel resistance would be ;

12-6.10= 5.90V

Now voltage drop in a parallel circuit is the same hence 5.90v is dropped in each of the parallel resistance.

That said, the current drop on the 10 ohm resistor would be;

5.90/10 = 0.59A

Remember V= I× R so that I = V/R

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(b)

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