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AVprozaik [17]
2 years ago
10

When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.3 a while

the second carries a current of 1.8
a?
Physics
1 answer:
stellarik [79]2 years ago
7 0

Correct question: When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.3 A while the second carries a current of 1.8 A .

What current will be supplied by the same battery if these two resistors are connected to it in series?

Answer:

1.164 A

Explanation:

From the question,

Let the unknown voltage of the batter = V.

Since the two resistors are connected in parallel,

The voltage across each of the resistor is the same = V,

Then,

From ohm's law,

V = IR................... Equation 1

Where V = Voltage, I = Current, R = Resistance.

make R the subject of the equation

R = V/I................ Equation 2

For the first resistor,

Given: I₁ = 3.3 A, V = V.

Substitute into equation 2

R₁ = V/3.3 Ω

For the second resistor,

Given: I₂ = 1.8 A

Substitute into equation 2

R₂ = V/1.8 Ω.

When the Resistors are connected in series,

Rt = R₁+R₂

Where Rt = Total resistance.

Rt = V/3.3+V/1.8

Rt = (1.8V+3.3V)/(3.3×1.8)

Rt = 5.1V/5.94

Rt = 0.859V.

Applying,

I' = V/Rt

where I' = current supplied by the battery, If the two resistors are connected in series

I' = V/0.859V

I' = 1.164 A

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1. Largest force: C;  smallest force: B; 2. ratio = 9:1

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Define forces exerted to the right as positive and those to the left as negative.

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\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

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