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?

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?

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

c. newton's third law of motion

a=d/t

a="triangle"v/t

explanation:

a = acceleration