Question

A 3-Ω light bulb and a 6-Ω light bulb are connected in series with a 6-volt battery. a) What is the current in each bulb? b) What is the voltage across each bulb? c) What is the power dissipated by each bulb? d) What is the total dissipated by the wire?

Answer 1

To be given in question

$R_1=3\Omega\\R_2=6\Omega\\V=6V$

$R_{eq}=R_1+R_2=3+6=9\Omega$

(A)

$Current$

$I=\frac{V}{R_{eq}}=\frac{6}{9}=0.66A$

(B)

Voltage

$V_1=I_1R_1=0.66\times3=1.98V\\V_2=I_2R_2=0.66\times6=3.96V$

(C)

$$P_1=I_1V_1=0.66\times1.98=1.3068W$$

$$P_2=I_2V_2=0.66\times3.96=2.6136W$$

(D) Total power

$$P=I_{t}\times R_{eq}=0.66\times9=5.94W$$