Can someone help me solve this its a simple series circuit

Indeed this is a serial circuit, in order to calculate the different questions, we must calculate the current of the circuit. The total resistance of the circuit can be calculated by means of an algebraic summation.

Then using ohm's law,we can calculate the current.

Rt = 2 + 4 + 6 = 12 [ohm]

The current

I = V / Rt

I = 12 / 12

I = 1 [amp]

Therefore the voltage in each resistance will be:

V1 = 2 * 1 = 2 [V]

V2 = 4 * 1 = 4 [V]

V3 = 6 * 1 = 6 [V]

The total voltage will be:

V = V1 + V2 + V3 = 12 [V]

The power can be calculated by multiplying the voltage by the total current.

P = V * I

P = 12 * 1

P = 12 [W]

Аноним Аноним · Answer 2 · 2021-08-26T07:57:19+02:00

Total Voltage will be 12V as only one battery connected

Now

[tex]\\ \sf\longmapsto R_{TOT}=R_1+R_2+R_3[/tex]

[tex]\\ \sf\longmapsto R_{TOT}=4+6+2[/tex]

[tex]\\ \sf\longmapsto R_{TOT}=12\Omega[/tex]

Lets find I

[tex]\\ \sf\longmapsto I_1=\dfrac{V}{R_1}=\dfrac{12}{2}=6A[/tex]

[tex]\\ \sf\longmapsto I_2=\dfrac{V}{R_2}=\dfrac{12}{4}=3A[/tex]

[tex]\\ \sf\longmapsto I_3=\dfrac{V}{R_3}=\dfrac{12}{6}=2A[/tex]

Hence

[tex]\\ \sf\longmapsto I_{Total}=6+3+2=11A[/tex]