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

hey , the answer is c

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards