Question

A-Some resistors are labelled with a red band. This shows that their true resistance will be within 2, point, 0, percent,2.0% of the value shown on the side. What is the largest current you would expect through a 15kΩ, 2.0% resistor when connected to a 5, point, 0, V,5.0V supply? Give your answer in m, A, mA.
B- When both of a car's red tail lamps are lit, this 'draws' a current of 0, point, 83, A,0.83A from the battery.Given that a car battery has a voltage of 12, V,12V, what is the resistance of each lamp? The lamps are wired in parallel.

Answer 1

A–The ratio of the voltage to the maximum resistor value gives the

largest current of 340.14 mA. In B–The lamp resistance is 28.9 Ω each.

Responses (approximate values):

A- The largest current is 340.14 mA

B- The resistance of each lamp is 28.9 Ω

Which methods can be used to find the current and resistances?

The range of values of the resistor = ±2%

The magnitude of the given resistor, R = 15 kΩ

The voltage supply, V = 5.0 V

Required:

Largest current, I, expected through the resistor.

Solution:

$Current, \, I = \mathbf{\dfrac{V}{R}}$

The current is largest when the resistor's value is smallest, which gives;

• $Largest \ current, \, I_{max} = \mathbf{ \dfrac{ 5.0 \, V}{\left(15 - \frac{2}{100} \times 15 \right) \Omega}} = \dfrac{50}{147} \, A \approx \underline{340.14 \, mA}$

B- Current drawn from car head lamps = 0.83 A

Voltage of the battery = 12 V

Required;

Resistance of each of the two lamps wired parallel.

Solution;

Let R represent the resistance of each lamp in the parallel connection,

we have;

$Equivalent \ resistance, \, R = \mathbf{ \dfrac{1}{\frac{1}{R} + \frac{1}{R} }} = \dfrac{R}{2}$

Therefore;

$\dfrac{R}{2} = \dfrac{12 \, V}{0.83 \, A}$

• $The \ resistance \ of \ each \ lamp, \, R = \dfrac{12 \, V}{0.83 \, A} \times 2 \approx \underline{28.9 \, \Omega}$

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