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Vikentia [17]
3 years ago
5

To practice problem-solving strategy 26.1 resistors in series and parallel. two bulbs are connected in parallel across a source

of emf e = 11.0 v with a negligible internal resistance. one bulb has a resistance of 3.0 Ï , and the other is 3.0 Ï . a resistor r is connected in the circuit in series with the two bulbs. what value of r should be chosen in order to supply each bulb with a voltage of 2.4 v ?
Physics
1 answer:
hichkok12 [17]3 years ago
3 0
Emf e = 11
r 1 = 3.0
r 2 = 3.0
r 3 = ?

The two in parallel are equivalent to 3 • 3/6 = 1.5 Ω 
To have 2.4 volts across them, the current is I = 2.4/1.5 = 1.6 amps. and the unknown R = (11–2.4) / 1.6 = 5.375 Ω or 5.4 Ω 
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Answer:

63.8 V

Explanation:

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We know that

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R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})

R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})

R=\rho l(18284.6)

I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}

Potential across 1.2 square cm=V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V

Hence, the voltage across the 1.2 square cm wire=63.8 V

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<span>==>  The total current is just the amount of current
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Diagram #2).
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