Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 months ago

# Three 15.0-W resistors are connected in parallel across a 30.0-V battery. What is the current through the battery?

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• 5 months ago

The resistance value is needed to solve for the current. the maximum dissipation of course is 45 watts.

If that is the power dissipated, the current would be 6 amps, R would be 15 ohm.

• 5 months ago

If the resistors are dissipating 15W each then:

15=E*I>>15/30= .5A

3*.5=1.5A

A fifteen watt resistor is a power specification of a resistor not the resistance of it. It means it can safely and continuously dissipate 15W.

If you mean 15 ohm resistors then:

Three 15 ohm resistors in parallel Have an Rt of 5 ohms.

I=E/R = 30/5 = 6A.

• oubaas
Lv 7
5 months ago

I = 15 watt @ 30 V = 0.5 A

It = I*3 = 1.5 A in total

• 5 months ago

The unit of resistance is ohms, not watts. Since the resistors are in parallel, the following equation is used to calculate the equivalent resistance.

1/Req = 1/R1 + 1/R2 + 1/R3

1.Req = 1/15 + 1/15 + 1/15 = 1/5

Req = 5 Ω

In parallel circuit, the voltage is the same for each resistor. Use the following equation to determine the current through each resistor.

I = V ÷ R

I1 = 30 ÷ 15 = 2 amperes

The total current is 6 amperes. I hope this is helpful for you.

• 5 months ago

The power rating of 15 W is the MAXIMUM HEAT that the resistor could handle.

The resistor is quite incapable of actually drawing 15W from any random power supply.

If they were ACTUALLY drawing 15W in this situation then the total power is 45 W

As power = v*i

i = P/V = 45 / 30 = 1.5 A total.

This is answering the question AS WRITTEN.

• 5 months ago

Are we sure that Hannah meant Ω instead of W?

'Cause W is usually Watts. Of course we would not be able to answer "What is the current?" if it IS intended to be Watts.

• 5 months ago

Resistors in parallel have the same voltage across them. Therefore

The total current leaving the source:

Itotal = 30/15 + 30/15 + 30/15 = 2 + 2 + 2 = 6A

OR find the resistance the 30V battery sees:

Req = 1/[1/15 + 1/15 + 1/ 15] = 1/[3/15] = 15/3 = 5Ω

• 5 months ago

V = I * R

1/R = 1/15 + 1/15 + 1/15

1/R = 3/15

R = 15/3

R = 5

V = I * R

30 = 5 * I

6 = I

6 amps