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2 years ago

5

A circuit diagram with a power source labeled 12 V is connected to 3 resistors in series. The resistors are labeled 1.0 Ohms, 2.

0 Ohms, and 3 Ohms. Calculate the equivalent resistance in the circuit shown.
0.55 Ω
1.8 Ω
2.0 Ω

Its A "0.55" I guessed and got it right

1 answer:

nevsk [136]2 years ago

7 0

Answer:

there it is fella u were right with ur answer

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What are the physical and chemical properties of sodium?

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Physical Properties of Sodium

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Home Periodic table Elements Sodium

Sodium - Na

Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium

Atomic number

11

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0.9

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1 year ago

The average lifetime of μ-mesons with a speed of 0.95c is measured to be 6 x 10^6 s. Find the average lifetime of μ-mesons in a

Mnenie [13.5K]

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19.2*10^6 s

Explanation:

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Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

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2 years ago

Nvm i got the answer but <br> free points

faust18 [17]

Answer:

thank you 谢谢

Explanation:

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2 years ago

Read 2 more answers

Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i

Lina20 [59]

Answer:

$214.9 \Omega$

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the $38.7 \Omega$ resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$

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2 years ago