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Soloha48 [4]
3 years ago
6

A light bulb has a resistance of 5 ohms and a maximum current of 10 A. How much voltage can be applied before the bulb will brea

k?​
Physics
1 answer:
Degger [83]3 years ago
8 0

The maximum voltage which can be applied before the bulb will break is 50 Volts.

<u>Given the following data:</u>

  • Resistance of light bulb = 5 Ohms
  • Current = 10 Amps.

To find how much voltage can be applied before the bulb will break, we would apply Ohm's law:

Ohm's law states that at constant temperature, the current flowing through an electrical circuit is directly proportional to the voltage applied across the two terminals of the circuit and it's inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

<u>Where;</u>

  • V is voltage measured in voltage.
  • I is current measured in amperes.
  • R is resistance measured in ohms.

Substituting the given parameters into the formula, we have;

Voltage = 10 × 5

<em>Voltage </em><em>=</em><em> 50 Volts.</em>

Therefore, the maximum voltage which can be applied before the bulb will break is 50 Volts.

Read more: brainly.com/question/23754122

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When you double your speed, it takes about _____ times as much distance to stop?
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A helium-filled balloon occupies a volume of 15 cubic meters at sea level. the balloon is released and raises to a point in the
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According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.

Therefore,

P_{1} V_{1} =P_{2} V_{2}

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5 0
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Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

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\Delta U = 1640 J

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Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

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Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
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Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
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