Bulbs in a motorcycle

In a motorcycle there is a battery with a voltage of 6 V. To the battery two devices are connected side by side. The first device is the bulb of the tail-light with an electrical resistance of 12 Ohm. The second is the bulb of the headlight with an electrical resistance of 1.0 Ohm.
a) Draw the circuit diagram described
b) Determine the current passing through the bulb of the tail-light
c) Determine the current passing through the bulb of the headlight
d) What current passes through the unbranched part of the circuit
e) Determine the resultant resistance of both bulbs

Final Answer:

I₁ =  0.5 A
I₂ =  6 A
I =  6.5 A
R =  0.9231 Ω

Step-by-step explanation:

$$\begin{gathered} U=6\text{V} \\ {R}_1=12\Omega \\ {R}_2=1.0\Omega \\ {I}_1=U/R_1=6/12=0.5\text{A} \end{gathered}$$

$I_2=U/R_2=6/1=6\text{A}$

$I=I_1+I_2=0.5+6=6.5\text{A}$

$R=U/I=6/6.5=0.9231\Omega$

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