Question

Three lightbulbs are connected to a 9-V battery as shown in Fig. 2.56 (a). Calculate:

(a) the total current supplied by the battery,

(b) the current through each bulb,

(c) the resistance of each bulb.

Answer 1

solution

(a) The total power supplied by the battery is equal to the total power absorbed by the bulbs; that is,

p=15+10+20=45W

Since p=VI, then the total current supplied by the battery is 

I=p/v=45/9=5A

(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since \( R_1 \)\( \)  (20-W bulb) is in parallel with the battery as well as the series combination of \( R_1 \) and \( R_2 \),

\( V_1=V_2+V_3 \)=9V

The current through \( R_1 \) is 

\( I_1=\frac{p_1}{V_1}=\frac{20}{9} \)=2.222A

By KCL, the current through the series combination of \( R_2 \) and \( R_3 \) is 

\( I_2=I-I_1 \)=5-2.222=2.778A

(c) the resistance of each bulb.

Since p=R\( I^2 \) =>\( R_1=\frac{p_1}{I_1^2}=\frac{20}{2.222^2}=4.05\Omega \)

                     =>\( R_2=\frac{p_1}{I_2^2}=\frac{15}{2.77^2}=1.945\Omega \)

                     =>\( R_3=\frac{p_3}{I_3^2}=\frac{10}{2.777^2}=1.297\Omega \)

 

Answer

(a) The total power supplied by the battery is equal to the total power absorbed by the bulbs; that is, I=5A

(b) The bulbs can be modeled as resistors as shown in Fig. 2.56(b). Since \( R_1 \)  (20-W bulb) is in parallel with the battery as well as the series combination of \( R_1 \) and \( R_2 \),

\( I_1 \)=2.222A and \( I_2 \)=2.788A

(c) \( R_1 \)=4.05\( \Omega \) , \( R_2 \)=1.945\( \Omega \) and \( R_3 \)=1.297\( \Omega \)