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ArbitrLikvidat [17]

2 years ago

5

A series circuit contains three 2 Ω resistors and a 6 Ω resistor. The current in each resistor is 4. 0 A. What is the voltage of

the battery in the circuit? 2. 0 V 3. 0 V 32 V 48 V.

1 answer:

Delvig [45]2 years ago

7 0

The voltage is a representation of the electric potential energy per unit charge.

The voltage across the circuit is 48 V. Option D shows the voltage across the circuit.

<h3>What is the voltage?</h3>

Voltage is defined as the difference in electric potential between two points.

Given that a series circuit contains three 2 Ω resistors and a 6 Ω resistor. The current in each resistor is 4.0 A.

The total resistance across the series circuit is given below.

$R = R_1 + R_2 + R_3 + R_4$

Where R is total resistance, R1, R2, R3 and R4 are resistors that are connected in series.

$R = 2+ 2+2+6$

$R = 12 \;\rm ohm$

The resistors are connected in series hence the current flows across the two points of the circuit are the same which is 4.0 A.

The voltage across the circuit is,

$V = IR$

$V = 4.0 \times 12$

$V = 48.0 \;\rm V$

Hence we can conclude that the voltage across the circuit is 48 V. Option D shows the voltage across the circuit.

To know more about the voltage, follow the link given below.

brainly.com/question/2364325.

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Given <u>rational inequality</u>:

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$\implies (x-1)^2=0\implies x=1$

$\implies (x-2)^3=0 \implies x=2$

Find the restrictions by solving f(x) = <em>undefined </em>(set the denominator to zero):

$\implies (x-3)^2(x-2)^2(x+5)=0$

$\implies (x-3)^2=0 \implies x=3$

$\implies (x-2)^2=0 \implies x=2$

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Create a sign chart, using closed dots for the <u>roots</u> and open dots for the <u>restrictions</u> (see attached).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

Test values: -6, 0, 1.5, 2.5, 4

For each test value, determine if the function is positive or negative:

$f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+$

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$f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

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Record the results on the sign chart for each region (see attached).

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