Collection of Solved Problems in Physics

Consumption of Car Light Bulbs

Task number: 1791

• Theory

  We need Ohm’s law for a segment of an electric circuit:

  \[ V = RI, \]

  where \(V\) is the voltage across a piece of conductor that has the resistance \(R\), and a current \(I\) flows through it.

  If there are resistors (such as light bulbs) connected in series, their resistances sum up

  \[ R_\mathrm{total} = R_1 + R_2 + \ldots + R_n, \]

  if they are connected in parallel, the reciprocals of the resistances sum up to the reciprocal of the total resistance

  \[ \frac{1}{R_\mathrm{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}, \] Considering the very low internal resistance of the car battery and the relatively low consumption of the light bulbs (compared to starting currents), we assume that the battery is an ideal source in this case, i.e. \(U \neq U(I)\).

• a) Hint – dipped headlights

  Draw the circuit diagram for the case when the dipped headlights are turned on. The bulbs are connected in parallel. How do we calculate the total resistance of such circuit?

  Calculate also the total current and the current through each light bulb.

• a) Hint solution – dipped headlights

  The following figure shows the circuit diagram for this situation.

  

  The diagram is simplified – the individual light circuits are protected by e.g. fuses. In reality, the switches control switching relays instead of directly switching the power lines of the respective light circuit branches.

  First, we calculate the total resistance. Two bulbs of dipped lights \(R_1\) and one bulb of the license plate lighting \(R_3\) are connected in parallel in this case.

  One can write for the total resistance \(R_\mathrm{t}\)

  \[ \frac{1}{R_\mathrm{t}}= \frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_3} = \frac{1}{2{.}5\,\mathrm{\Omega}}+ \frac{1}{2{.}5\,\mathrm{\Omega}}+ \frac{1}{30\,\mathrm{\Omega}} = \frac{1}{1{.}2\,\mathrm{\Omega}}, \]

  and then

  \[ R_\mathrm{t}= 1{.}2\,\mathrm{\Omega}. \]

  The total current is

  \[ I = \frac{V}{R_\mathrm{t}} = \frac{12\,\mathrm{V}}{1{.}2\,\mathrm{\Omega}}= 10\,\mathrm{A}. \]

  All bulbs are connected to the battery with voltage \(V\). The current through one dipped light bulb is simply

  \[ I_1 = \frac{V}{R_1} = \frac{12\,\mathrm{V}}{2{.}5\,\mathrm{\Omega}} = 4{.}8\,\mathrm{A}, \]

  and the current through the licenste plate light bulb is

  \[ I_3 = \frac{V}{R_3} = \frac{12\,\mathrm{V}}{30\,\mathrm{\Omega}} = 0{.}4\,\mathrm{A}. \]

  We check that by summing of the currents drawn by all the bulbs

  \[ I = 2I_1 + I_3 = 2{\cdot} 4{.}8\,\mathrm{A} + 0{.}4\,\mathrm{A} = 10\,\mathrm{A} \]

    we indeed get the value of the total current.

• b) Hint – Switching on the fog lights

  Draw the circuit diagram for the case when the fog lights are turned on in addition to the previous case. The bulbs are connected in parallel again. How do we calculate the total resistance of the circuit?

  Calculate also the total current and the current through respective light bulbs.

• b) Hint solution – Switching on the fog lights

  The following figure shows the circuit diagram for this situation.

  

  The diagram is simplified – the individual light circuits are protected by e.g. fuses. In reality, the switches control switching relays instead of directly switching the power lines of the respective light circuit branches.

  We first calculate the total resistance. Two bulbs of dipped lights \(R_1\), two bulbs of fog lights \(R_2\) and one bulb of the license plate lighting \(R_3\) are connected in parallel in this case.

  One can write for the total resistance \(R_\mathrm{t}\)

  \[ \frac{1}{R_\mathrm{t}}= \frac{1}{R_1}+\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_2}+\frac{1}{R_3} = \] \[ = \frac{1}{2{.}5\,\mathrm{\Omega}}+ \frac{1}{2{.}5\,\mathrm{\Omega}}+ \frac{1}{4\,\mathrm{\Omega}}+ \frac{1}{4\,\mathrm{\Omega}}+ \frac{1}{30\,\mathrm{\Omega}} = \frac{1}{0{.}75\,\mathrm{\Omega}}, \]

  and then

  \[ R_\mathrm{t}= 0{.}75\,\mathrm{\Omega}. \]

  The total current through the circuit is

  \[ I = \frac{U}{R_\mathrm{t}} = \frac{12\,\mathrm{V}}{0{.}75\,\mathrm{\Omega}}= 16\,\mathrm{A}. \]

  All bulbs are connected to the battery with voltage \(V\). The current through one dipped light bulb is again

  \[ I_1 = \frac{U}{R_1} = \frac{12\,\mathrm{V}}{2{.}5\,\mathrm{\Omega}} = 4{.}8\,\mathrm{A}, \]

  the current through the licenste plate light bulb also remains the same

  \[ I_3 = \frac{U}{R_3} = \frac{12\,\mathrm{V}}{30\,\mathrm{\Omega}} = 0{.}4\,\mathrm{A}. \] One fog light bulb draws a current \[ I_2 = \frac{U}{R_2} = \frac{12\,\mathrm{V}}{4\,\mathrm{\Omega}} = 3\,\mathrm{A}. \]

  We check that by summingof the currents drawn by all the bulbs

  \[ I = 2I_1 + 2I_2 + I_3 = 2{\cdot} 4{.}8\,\mathrm{A} + 2{\cdot} 3\,\mathrm{A} + 0{.}4\,\mathrm{A} = 16\,\mathrm{A} \]

    we indeed get the value of the total current.

• Answer

  The total resistance of the circuit is \(1{.}2\,\mathrm{\Omega}\) when only the dipped lights with the license plate lighting are turned on. The resistance of the circuit decreases to \(0{.}75\,\mathrm{\Omega}\) when we turn on the fog lights in addition.

  The current drawn by the lighting circuit increases from \(10\,\mathrm{A}\) to \(16\,\mathrm{A}\) after turning on the fog lights.

  One dipped light bulb draws a current of \(4{.}8\,\mathrm{A}\), the fog light bulb draws \(3\,\mathrm{A}\) and the license plate light bulb draws \(0{.}4\,\mathrm{A}\).