Question

Some makes of older cars have \(6.00 - V\)electrical systems.

(a) What is the hot resistance of a \(30.0 - W\)headlight in such a car?

(b) What current flows through it?

Short answer

1. The hot resistance of the headlight of the car is \(1.20{\rm{ }}\Omega \) .
2. The current flow through the headlight of the car is \(5.00\;A\).

Step-by-step solution

Ohm's law

The power is defined by the formula

\(\begin{align}{}P &= I\Delta V\\ &= \frac{{{{\left( {\Delta V} \right)}^2}}}{R}......(2)\end{align}\)

Where\(I\)is the current and\(\Delta V\)is potential difference.

Definition of Power

The power is defined by the formula

\(\begin{align}{}P &= I\Delta V\\ &= \frac{{{{\left( {\Delta V} \right)}^2}}}{R}......(2)\end{align}\)

Where\(I\)is the current and\(\Delta V\)is potential difference.

The given data

• The potential difference across the electrical system is:\(\Delta V = 6.00\;V\).
• The power consumption of the headlight is: \(P = 30.0\;W\).

Calculation of the resistance

a. The resistance of the headlight is found by solving Equation \((2)\) for \(R\):\(R = \frac{{{{(\Delta V)}^2}}}{P}\)

Substituting the values for \(\Delta V\)and \(P\)in the above expression, and we get,

\(\begin{align}{}R &= \frac{{{{(6.00\;V)}^2}}}{{30.0\;W}}\\ &= 1.20{\rm{ }}\Omega \end{align}\)

Therefore, the resistance is \(1.20{\rm{ }}\Omega \).

Calculation for the current

b. The current in the headlight is found by solving Equation \((1)\) for \(I\) :

\(I = \frac{P}{{\Delta V}}\)

Substituting the values for \(P\) and \(\Delta V\) in the above expression, and we get,

\(\begin{align}{}I &= \frac{{30.0\;W}}{{6.00\;V}}\\ &= 5.00\;A\end{align}\)

Therefore, the current flow is \(5.00\;A\).